• ccpc 2016 changchun 长春（4.12训练）

    概述

    训练来源： ccpc2016 长春赛区区域赛

    训练时间： 2019-04-02 13:00 至 2019-04-02 18:00

    训练人： jmx，cy

    通过题目： 7/12：B,D,F,G,H,I,J

    排名： 11/181

    金牌题数： 7

    出线题数： 9

    赛后补题： E,K

    题解(按照题目通过顺序)

    B - Fraction (cy)

    题目来源： HDU - 5912

    00:07:54 通过 (1次提交)

    签到题，cy码的比较稳

    #include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <map>
#include <complex>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <bitset>
#include <cmath>
#include <set>

int N = 1e6, SZ = 320, INF = 1 << 29;
long long LINF = (1LL << 61), mod = 1e9 + 7;
const long double eps = 1e-9, PI = acos(-1.0);

#define lowbit(x) (x & (-(x)))
#define MAX(a, b) ((a) < (b) ? (b) : (a))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define rp(a, b, c) for (int a = b; a <= c; ++a)
#define RP(a, b, c) for (int a = b; a < c; ++a)
#define lp(a, b, c) for (int a = b; a >= c; --a)
#define LP(a, b, c) for (int a = b; a > c; --a)
#define rps(i, s) for (int i = 0; s[i]; i++)
#define fson(u) for (int i = g[u]; ~i; i = edg[i].nxt)
#define adde(u, v) edg[++ecnt] = Edge(u, v, 0, g[u]), g[u] = ecnt
#define addew(u, v, w) edg[++ecnt] = Edge(u, v, w, g[u]), g[u] = ecnt
#define MID (l + r >> 1)
#define mst(a, v) memset(a, v, sizeof(a))
#define bg(x)            
	Edge edg[maxn << x]; 
	int g[maxn], ecnt
#define ex(v)  
	cout << v; 
	return 0
#define debug(x) cout << "debug: " << x << endl;
#define sqr(x) ((x) * (x))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef complex<double> cpx;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef map<int, int> mii;
typedef map<ll, ll> mll;

char READ_DATA;
int SIGNAL_INPUT;
template <typename Type>
inline Type ru(Type &v)
{
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
		v = v * 10 + READ_DATA - '0';
	v *= SIGNAL_INPUT;
	return v;
}
inline ll modru(ll &v)
{
	ll p = 0;
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	p = v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
	{
		v = (v * 10 + READ_DATA - '0') % mod;
		p = (p * 10 + READ_DATA - '0') % (mod - 1);
	}
	v *= SIGNAL_INPUT;
	return p;
}
template <typename A, typename B>
inline int ru(A &x, B &y)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	return 2;
}
template <typename A, typename B, typename C>
inline int ru(A &x, B &y, C &z)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	return 3;
}
template <typename A, typename B, typename C, typename D>
inline int ru(A &x, B &y, C &z, D &w)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	ru(w);
	return 4;
}
inline ll gcd(ll a, ll b)
{
	while (b)
	{
		a %= b;
		swap(a, b);
	}
	return a;
}

inline ll fastmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return (a * b - (ll)((long double)a * b / mod) * mod + mod) % mod;
}

inline ll dirmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return a * b % mod;
}

inline ll ss(ll a, ll b, ll mod = 1e9 + 7, ll(*mul)(ll, ll, ll) = dirmul)
{
	if (b < 0)
	{
		b = -b;
		a = ss(a, mod - 2, mod);
	}
	ll ans = 1;
	while (b)
	{
		if (b & 1)
			ans = mul(ans, a, mod);
		a = mul(a, a, mod);
		b >>= 1;
	}
	return ans;
}
inline int isprime(ll n)
{
	if (n == 1)
		return 0;

	for (ll d = 2; d * d <= n; ++d)
	{
		if (n % d == 0)
			return 0;
	}

	return 1;
}

template <typename Type>
void brc(Type *a, int n)
{
	int k;
	for (int i = 1, j = n / 2; i < n - 1; i++)
	{
		if (i < j)
			swap(a[i], a[j]);

		k = n >> 1;
		while (j >= k)
		{
			j ^= k;
			k >>= 1;
		}
		if (j < k)
			j ^= k;
	}
}
void fft(cpx *a, int n, int inv = 1)
{
	cpx u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		cpx wn(cos(inv * 2.0 * PI / h), sin(inv * 2.0 * PI / h));
		for (int j = 0; j < n; j += h)
		{
			cpx w(1, 0);
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)];
				a[k] = u + t;
				a[k + (h >> 1)] = u - t;
				w *= wn;
			}
		}
	}
	if (inv == -1)
		RP(i, 0, n)
		a[i] /= n;
}
void ntt(ll *a, int n, int inv = 1)
{
	ll u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		ll wn = ss(3, (mod - 1) / h);
		if (inv == -1)
			wn = ss(wn, mod - 2);
		for (int j = 0; j < n; j += h)
		{
			ll w = 1;
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)] % mod;
				a[k] = (u + t) % mod;
				a[k + (h >> 1)] = (u - t + mod) % mod;
				(w *= wn) %= mod;
			}
		}
	}
	if (inv == -1)
	{
		ll tmp = ss(n, mod - 2);
		RP(i, 0, n)
			(a[i] *= tmp) %= mod;
	}
}
struct Edge
{
	int u, v, nxt;
	//ll w;
	Edge(int _u = 0, int _v = 0, /*ll _w = 0,*/ int _nxt = 0)
	{
		u = _u;
		v = _v;
		//w = _w;
		nxt = _nxt;
	}

	/*int operator<(const Edge &b) const
	{
		return w < b.w;
	}*/
};
struct CMP
{
	int operator()(const int &a, const int &b) const
	{
		return a > b;
	}
};

const int maxn = 1e2+ 5;
/*------------------------------------------------------------------------yah01------------------------------------------------------------------------*/

int n, a[maxn], b[maxn];

pll f(int i)
{
	if (i == n)
	{
		ll d = gcd(a[n], b[n]);
		return pll(b[n] / d, a[n] / d);
	}

	pll t = f(i + 1);
	ll x = b[i] * t.second, y = a[i] * t.second + t.first;
	ll d = gcd(x, y);
	return pll(x / d, y / d);
}

int main()
{	
	int T;
	ru(T);
	rp(t,1,T)
	{
		ru(n);
		rp(i, 1, n) ru(a[i]);
		rp(i, 1, n) ru(b[i]);

		pll ans = f(1);
		printf("Case #%d: %lld %lld
", t,ans.first, ans.second);
	}
	return 0;
}

/*
5
2 4 6 8 10
-1 2 2 2
1000
+ 1 10
+ 1 -2
+ 1 1
*/

    D - Triangle (jmx)

    题目来源： HDU-5914

    00:19:22 通过（一次提交）

    暴力枚举即可。签到题。

    #include <bits/stdc++.h>
#define ll long long
#define inf 0x6fffffff
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int mx=(1<<20);
int f[30],a[30];
int main()
{
    for(int op=0;op<mx;op++)
    {
        int cnt=0;
        for(int i=0;i<20;i++)
            if(((1<<i)&op)>0)a[++cnt]=i+1;
        if(cnt<=2)
        {
            for(int i=a[cnt];i<=20;i++)f[i]=max(cnt,f[i]);
        }
        else{
            bool flag=0;
            for(int i=1;i<=cnt-2;i++)
            {
                if(a[i]+a[i+1]>a[i+2]){flag=1;break;}
            }
            if(!flag)
                for(int i=a[cnt];i<=20;i++)f[i]=max(cnt,f[i]);
        }
    }
    for(int i=1;i<=20;i++)f[i]=i-f[i];
    int T=read();
    for(int cas=1;cas<=T;cas++)
    {
        int n=read();
        printf("Case #%d: %d
",cas,f[n]);
    }
    return 0;
}

    F - Harmonic Value Description (jmx)

    题目来源：HDU-5916

    00:31:04 通过（1次提交）

    通过观察可以得出:

    1. $P_i,P_{i-1}$一个是奇数一个是偶数时，$gcd(P_i,P_{i-1})=1$
    2. $P_i,P_{i-1}$为连续的奇数时，$gcd(P_i,P_{i-1})=1$
    3. $P_i,P_{i-1}$为连续的偶数时，$gcd(P_i,P_{i-1})=2$

    那么一定可以通过如下方法构造出所求和为$n-1+i$的序列：将前$i$个偶数按顺序放在最前面，然后将前$i$个奇数按顺序放在这些偶数的后面，再将剩下的所有数按顺序放在最后。

    #include <bits/stdc++.h>
#define ll long long
#define inf 0x6fffffff
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
const int mx=(1<<20);
int f[30],a[30];
int main()
{
    for(int op=0;op<mx;op++)
    {
        int cnt=0;
        for(int i=0;i<20;i++)
            if(((1<<i)&op)>0)a[++cnt]=i+1;
        if(cnt<=2)
        {
            for(int i=a[cnt];i<=20;i++)f[i]=max(cnt,f[i]);
        }
        else{
            bool flag=0;
            for(int i=1;i<=cnt-2;i++)
            {
                if(a[i]+a[i+1]>a[i+2]){flag=1;break;}
            }
            if(!flag)
                for(int i=a[cnt];i<=20;i++)f[i]=max(cnt,f[i]);
        }
    }
    for(int i=1;i<=20;i++)f[i]=i-f[i];
    int T=read();
    for(int cas=1;cas<=T;cas++)
    {
        int n=read();
        printf("Case #%d: %d
",cas,f[n]);
    }
    return 0;
}

    H - Sequence I (cy)

    题目来源：HDU-5918

    01:03:06 通过（1次提交）

    根据题意暴力枚举匹配，使用kmp算法加速即可

    #include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <map>
#include <complex>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <bitset>
#include <cmath>
#include <set>

int N = 1e6, SZ = 320, INF = 1 << 29;
long long LINF = (1LL << 61), mod = 1e9 + 7;
const long double eps = 1e-9, PI = acos(-1.0);

#define lowbit(x) (x & (-(x)))
#define MAX(a, b) ((a) < (b) ? (b) : (a))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define rp(a, b, c) for (int a = b; a <= c; ++a)
#define RP(a, b, c) for (int a = b; a < c; ++a)
#define lp(a, b, c) for (int a = b; a >= c; --a)
#define LP(a, b, c) for (int a = b; a > c; --a)
#define rps(i, s) for (int i = 0; s[i]; i++)
#define fson(u) for (int i = g[u]; ~i; i = edg[i].nxt)
#define adde(u, v) edg[++ecnt] = Edge(u, v, 0, g[u]), g[u] = ecnt
#define addew(u, v, w) edg[++ecnt] = Edge(u, v, w, g[u]), g[u] = ecnt
#define MID (l + r >> 1)
#define mst(a, v) memset(a, v, sizeof(a))
#define bg(x)            
	Edge edg[maxn << x]; 
	int g[maxn], ecnt
#define ex(v)  
	cout << v; 
	return 0
#define debug(x) cout << "debug: " << x << endl;
#define sqr(x) ((x) * (x))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef complex<double> cpx;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef map<int, int> mii;
typedef map<ll, ll> mll;

char READ_DATA;
int SIGNAL_INPUT;
template <typename Type>
inline Type ru(Type &v)
{
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
		v = v * 10 + READ_DATA - '0';
	v *= SIGNAL_INPUT;
	return v;
}
inline ll modru(ll &v)
{
	ll p = 0;
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	p = v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
	{
		v = (v * 10 + READ_DATA - '0') % mod;
		p = (p * 10 + READ_DATA - '0') % (mod - 1);
	}
	v *= SIGNAL_INPUT;
	return p;
}
template <typename A, typename B>
inline int ru(A &x, B &y)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	return 2;
}
template <typename A, typename B, typename C>
inline int ru(A &x, B &y, C &z)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	return 3;
}
template <typename A, typename B, typename C, typename D>
inline int ru(A &x, B &y, C &z, D &w)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	ru(w);
	return 4;
}
inline ll gcd(ll a, ll b)
{
	while (b)
	{
		a %= b;
		swap(a, b);
	}
	return a;
}

inline ll fastmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return (a * b - (ll)((long double)a * b / mod) * mod + mod) % mod;
}

inline ll dirmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return a * b % mod;
}

inline ll ss(ll a, ll b, ll mod = 1e9 + 7, ll(*mul)(ll, ll, ll) = dirmul)
{
	if (b < 0)
	{
		b = -b;
		a = ss(a, mod - 2, mod);
	}
	ll ans = 1;
	while (b)
	{
		if (b & 1)
			ans = mul(ans, a, mod);
		a = mul(a, a, mod);
		b >>= 1;
	}
	return ans;
}
inline int isprime(ll n)
{
	if (n == 1)
		return 0;

	for (ll d = 2; d * d <= n; ++d)
	{
		if (n % d == 0)
			return 0;
	}

	return 1;
}

template <typename Type>
void brc(Type *a, int n)
{
	int k;
	for (int i = 1, j = n / 2; i < n - 1; i++)
	{
		if (i < j)
			swap(a[i], a[j]);

		k = n >> 1;
		while (j >= k)
		{
			j ^= k;
			k >>= 1;
		}
		if (j < k)
			j ^= k;
	}
}
void fft(cpx *a, int n, int inv = 1)
{
	cpx u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		cpx wn(cos(inv * 2.0 * PI / h), sin(inv * 2.0 * PI / h));
		for (int j = 0; j < n; j += h)
		{
			cpx w(1, 0);
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)];
				a[k] = u + t;
				a[k + (h >> 1)] = u - t;
				w *= wn;
			}
		}
	}
	if (inv == -1)
		RP(i, 0, n)
		a[i] /= n;
}
void ntt(ll *a, int n, int inv = 1)
{
	ll u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		ll wn = ss(3, (mod - 1) / h);
		if (inv == -1)
			wn = ss(wn, mod - 2);
		for (int j = 0; j < n; j += h)
		{
			ll w = 1;
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)] % mod;
				a[k] = (u + t) % mod;
				a[k + (h >> 1)] = (u - t + mod) % mod;
				(w *= wn) %= mod;
			}
		}
	}
	if (inv == -1)
	{
		ll tmp = ss(n, mod - 2);
		RP(i, 0, n)
			(a[i] *= tmp) %= mod;
	}
}
struct Edge
{
	int u, v, nxt;
	//ll w;
	Edge(int _u = 0, int _v = 0, /*ll _w = 0,*/ int _nxt = 0)
	{
		u = _u;
		v = _v;
		//w = _w;
		nxt = _nxt;
	}

	/*int operator<(const Edge &b) const
	{
		return w < b.w;
	}*/
};
struct CMP
{
	int operator()(const int &a, const int &b) const
	{
		return a > b;
	}
};

const int maxn = 1e6+ 5;
/*------------------------------------------------------------------------yah01------------------------------------------------------------------------*/

int n, m, p;
int a[maxn], b[maxn];
int f[maxn];

void KMP()
{
	int j;
	f[0] = f[1] = 0;
	RP(i, 1, m)
	{
		j = f[i];
		while (j && b[i] != b[j]) j = f[j];
		f[i + 1] = b[i] == b[j] ? j + 1 : 0;
	}
}

int main()
{	
	int T;
	ru(T);
	rp(t,1,T)
	{
		ru(n, m, p);
		RP(i, 0, n) ru(a[i]);
		RP(i, 0, m) ru(b[i]);
		KMP();

		int j, ans = 0;
		RP(s, 0, p)
		{
			j = 0;
			for (int i = s; i < n; i += p)
			{
				while (j && a[i] != b[j]) j = f[j];
				if (a[i] == b[j]) ++j;
				if (j == m)
				{
					++ans;
					j = f[j];
				}
			}
		}

		printf("Case #%d: %d
", t, ans);
	}
	return 0;
}

/*
5
2 4 6 8 10
-1 2 2 2
1000
+ 1 10
+ 1 -2
+ 1 1
*/

    I - Sequence II (jmx)

    题目来源：HDU-5919

    02:03:53 通过（2次提交）

    统计$[l,r]$之间只出现一次的数的个数，我们利用主席树记录相同的数的前一个位置是多少，很容易得到区间有多少个不同的数。然后可以通过二分来解决，但是$O(nlogn)$的时间复杂度怕超时就没写。

    考虑一下可以倒着插入，某个数出现的位置$x$，上一次出现的位置为$y$，则在$root_x$处的$x$位置+1，在$root_y$处的$x$位置-1。求一个后缀和，$root_l$中记录的就是从所有以$l$为左端点的区间中有多少个位置的数是第一次出现的。对于每次询问，我们只要在$root_l$的线段树上先求出$[l,r]$有多少个数（假设为t），并在$root_l$的线段树上二分出第$lfloor (t+1)/2 floor$个数的位置就行了。

    #include <bits/stdc++.h>
#define ll long long
#define inf 0x6fffffff
#define N 200086
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
map<int,int>mp;
int a[N],pre[N],nxt[N],root[N];
int n,m;
int cnt;
struct node{int l,r,w;}f[N*80];
void ins(int &i,int l,int r,int ps,int dd)
{
    f[++cnt]=f[i];i=cnt;f[cnt].w+=dd;
    if(l==r)return;
    int mid=(l+r)/2;
    if(ps<=mid)ins(f[i].l,l,mid,ps,dd);else ins(f[i].r,mid+1,r,ps,dd);
}
int ask(int i,int l,int r,int lt,int rt)
{
    if(lt<=l&&r<=rt)return f[i].w;
    int mid=(l+r)/2,res=0;
    if(lt<=mid)res+=ask(f[i].l,l,mid,lt,rt);
    if(mid+1<=rt)res+=ask(f[i].r,mid+1,r,lt,rt);
    return res;
}
int find(int i,int l,int r,int rk)
{
    if(l==r)return l;
    int mid=(l+r)/2,tmp=f[f[i].l].w;
    if(tmp>=rk)return find(f[i].l,l,mid,rk);
        else return find(f[i].r,mid+1,r,rk-tmp);
    
}
int main()
{
    int T=read();
    for(int cas=1;cas<=T;cas++)
    {
        int n=read(),m=read();
        printf("Case #%d:",cas);
        cnt=0;mp.clear();
        for(int i=1;i<=n+1;i++)root[i]=pre[i]=nxt[i]=0;
        for(int i=1;i<=n;i++)
        {
            a[i]=read();
            pre[i]=mp[a[i]];mp[a[i]]=i;
            if(pre[i]!=0)nxt[pre[i]]=i;
        }
        for(int i=n;i;i--)
        {
            root[i]=root[i+1];
            ins(root[i],1,n,i,1);
            if(nxt[i]!=0)ins(root[i],1,n,nxt[i],-1);
        }
        int ans=0;
        for(int i=1;i<=m;i++)
        {
            int l=(read()+ans)%n+1,r=(read()+ans)%n+1;
            if(l>r)swap(l,r);
            int k=ask(root[l],1,n,l,r);
            int rk=(k+1)/2;
            ans=find(root[l],1,n,rk);
            printf(" %d",ans);
        }
        puts("");
    }
    return 0;
}

    J - Ugly Problem (cy)

    题目来源：HDU-5920

    02:49:50 通过（5次提交）

    将数字的前一半提取出来，然后将这个数字减$1$，然后把这个数字做对称形成后一半。用原来的数减去这个小于原数的回文数，得到一个新的数。之后对新的数执行以上操作，直到变成一位数。这样可以保证每次都可以把数字的长度减小一半。

    #include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <map>
#include <complex>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <bitset>
#include <cmath>
#include <set>

int N = 1e6, SZ = 320, INF = 1 << 29;
long long LINF = (1LL << 61), mod = 1e9 + 7;
const long double eps = 1e-9, PI = acos(-1.0);

#define lowbit(x) (x & (-(x)))
#define MAX(a, b) ((a) < (b) ? (b) : (a))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define rp(a, b, c) for (int a = b; a <= c; ++a)
#define RP(a, b, c) for (int a = b; a < c; ++a)
#define lp(a, b, c) for (int a = b; a >= c; --a)
#define LP(a, b, c) for (int a = b; a > c; --a)
#define rps(i, s) for (int i = 0; s[i]; i++)
#define fson(u) for (int i = g[u]; ~i; i = edg[i].nxt)
#define adde(u, v) edg[++ecnt] = Edge(u, v, 0, g[u]), g[u] = ecnt
#define addew(u, v, w) edg[++ecnt] = Edge(u, v, w, g[u]), g[u] = ecnt
#define MID (l + r >> 1)
#define mst(a, v) memset(a, v, sizeof(a))
#define bg(x)            
	Edge edg[maxn << x]; 
	int g[maxn], ecnt
#define ex(v)  
	cout << v; 
	return 0
#define debug(x) cout << "debug: " << x << endl;
#define sqr(x) ((x) * (x))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef complex<double> cpx;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef map<int, int> mii;
typedef map<ll, ll> mll;

char READ_DATA;
int SIGNAL_INPUT;
template <typename Type>
inline Type ru(Type &v)
{
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
		v = v * 10 + READ_DATA - '0';
	v *= SIGNAL_INPUT;
	return v;
}
inline ll modru(ll &v)
{
	ll p = 0;
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	p = v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
	{
		v = (v * 10 + READ_DATA - '0') % mod;
		p = (p * 10 + READ_DATA - '0') % (mod - 1);
	}
	v *= SIGNAL_INPUT;
	return p;
}
template <typename A, typename B>
inline int ru(A &x, B &y)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	return 2;
}
template <typename A, typename B, typename C>
inline int ru(A &x, B &y, C &z)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	return 3;
}
template <typename A, typename B, typename C, typename D>
inline int ru(A &x, B &y, C &z, D &w)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	ru(w);
	return 4;
}
inline ll gcd(ll a, ll b)
{
	while (b)
	{
		a %= b;
		swap(a, b);
	}
	return a;
}

inline ll fastmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return (a * b - (ll)((long double)a * b / mod) * mod + mod) % mod;
}

inline ll dirmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return a * b % mod;
}

inline ll ss(ll a, ll b, ll mod = 1e9 + 7, ll(*mul)(ll, ll, ll) = dirmul)
{
	if (b < 0)
	{
		b = -b;
		a = ss(a, mod - 2, mod);
	}
	ll ans = 1;
	while (b)
	{
		if (b & 1)
			ans = mul(ans, a, mod);
		a = mul(a, a, mod);
		b >>= 1;
	}
	return ans;
}
inline int isprime(ll n)
{
	if (n == 1)
		return 0;

	for (ll d = 2; d * d <= n; ++d)
	{
		if (n % d == 0)
			return 0;
	}

	return 1;
}

template <typename Type>
void brc(Type *a, int n)
{
	int k;
	for (int i = 1, j = n / 2; i < n - 1; i++)
	{
		if (i < j)
			swap(a[i], a[j]);

		k = n >> 1;
		while (j >= k)
		{
			j ^= k;
			k >>= 1;
		}
		if (j < k)
			j ^= k;
	}
}
void fft(cpx *a, int n, int inv = 1)
{
	cpx u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		cpx wn(cos(inv * 2.0 * PI / h), sin(inv * 2.0 * PI / h));
		for (int j = 0; j < n; j += h)
		{
			cpx w(1, 0);
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)];
				a[k] = u + t;
				a[k + (h >> 1)] = u - t;
				w *= wn;
			}
		}
	}
	if (inv == -1)
		RP(i, 0, n)
		a[i] /= n;
}
void ntt(ll *a, int n, int inv = 1)
{
	ll u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		ll wn = ss(3, (mod - 1) / h);
		if (inv == -1)
			wn = ss(wn, mod - 2);
		for (int j = 0; j < n; j += h)
		{
			ll w = 1;
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)] % mod;
				a[k] = (u + t) % mod;
				a[k + (h >> 1)] = (u - t + mod) % mod;
				(w *= wn) %= mod;
			}
		}
	}
	if (inv == -1)
	{
		ll tmp = ss(n, mod - 2);
		RP(i, 0, n)
			(a[i] *= tmp) %= mod;
	}
}
struct Edge
{
	int u, v, nxt;
	//ll w;
	Edge(int _u = 0, int _v = 0, /*ll _w = 0,*/ int _nxt = 0)
	{
		u = _u;
		v = _v;
		//w = _w;
		nxt = _nxt;
	}

	/*int operator<(const Edge &b) const
	{
		return w < b.w;
	}*/
};
struct CMP
{
	int operator()(const int &a, const int &b) const
	{
		return a > b;
	}
};

const int maxn = 1e3 + 5;
/*------------------------------------------------------------------------yah01------------------------------------------------------------------------*/

int n;
char s[maxn];
vector<string> ans;
int main()
{
	int T;
	ru(T);
	rp(t, 1, T)
	{
		//srand(47);
		scanf("%s", s + 1);
		n = strlen(s + 1);
		//n = 1000;
		//rp(i, 2, n) s[i] = rand() % 10 + '0';
		//s[1] = rand() % 9 + 1 + '0';
		reverse(s + 1, s + n + 1);
		rp(i, 1, n)
		{
			s[i] = s[i] - '0';
		}
		printf("Case #%d:
", t);

		ans.clear();
		string now;
		int x, cnt = 0;
		while (n)
		{
			int m = n / 2;
			int choice = 0;
			if (1)
			{
				if (n>=2 && s[n / 2 + 1] == 0)
				{
					now.clear();
					rp(i, 1, n / 2 + 1) now.push_back('1');
				}
				else
				{
					rp(i, 1, m)
					{
						if (s[i] < s[n - i + 1]) break;
						else if (s[i] > s[n - i + 1])
						{
							choice = 1;
							break;
						}
					}

					now.clear();
					choice = 0;
					if (n & 1)
					{
						if (n == 1)
						{
							now.push_back(s[1]+'0');
						}
						else
						{
							rp(i, 1, m) now.push_back(s[n - i + 1] + '0');
							now.push_back(s[n / 2 + 1] - 1 + '0');
							lp(i, m - 1, 0) now.push_back(now[i]);
						}
					}
					else
					{
						if (n == 2)
						{
							int x = min(s[1], s[2]);
							if (x)
							{
								now.push_back(x + '0');
								now.push_back(x + '0');
							}
							else
							{
								now.push_back('1');
							}
						}
						else
						{
							rp(i, 1, m - 1) now.push_back(s[n-i+1] + '0');
							now.push_back(s[m + 1] - 1 + '0');
							//now.push_back(s[m + 1] - 1 + '0');
							lp(i, m-1, 0) now.push_back(now[i]);
						}
					}
					
				}
			}
			else
			{
				now.clear();
				now.push_back(10 - s[n] + '0');
			}

			rp(i, 1, n)
			{
				if(i-1<now.length())
					s[i] -= now[i - 1] - '0';
				while (s[i] < 0)
				{
					s[i] += 10;
					s[i + 1]--;
				}
			}

			while (n&&s[n] == 0) n--;
			ans.push_back(now);
		}
		
		int tot = ans.size();
		printf("%d
", tot);
		RP(i, 0, tot)
		{
			int m = ans[i].length();
			RP(j, 0, m) printf("%c", ans[i][j]);
			putchar('
');
		}
	}
	return 0;
}

/*
5
2 4 6 8 10
-1 2 2 2
1000
+ 1 10
+ 1 -2
+ 1 1
*/

    G - Instability (cy&jmx)

    题目来源：HDU-5917

    03:28:07 通过（3次提交）

    根据Ramsey定理，任意包含不少于$6$个点的集合，总会至少存在一个大小为$3$的团或者一个大小为$3$的独立集。

    这样的话只要暴力枚举小于$6$个点的集合的稳定性就行了，时间复杂度完全够用。

    #include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <vector>
#include <map>
#include <complex>
#include <queue>
#include <algorithm>
#include <string>
#include <stack>
#include <bitset>
#include <cmath>
#include <set>

int N = 1e6, SZ = 320, INF = 1 << 29;
long long LINF = (1LL << 61), mod = 1e9 + 7;
const long double eps = 1e-9, PI = acos(-1.0);

#define lowbit(x) (x & (-(x)))
#define MAX(a, b) ((a) < (b) ? (b) : (a))
#define MIN(a, b) ((a) < (b) ? (a) : (b))
#define rp(a, b, c) for (int a = b; a <= c; ++a)
#define RP(a, b, c) for (int a = b; a < c; ++a)
#define lp(a, b, c) for (int a = b; a >= c; --a)
#define LP(a, b, c) for (int a = b; a > c; --a)
#define rps(i, s) for (int i = 0; s[i]; i++)
#define fson(u) for (int i = g[u]; ~i; i = edg[i].nxt)
#define adde(u, v) edg[++ecnt] = Edge(u, v, 0, g[u]), g[u] = ecnt
#define addew(u, v, w) edg[++ecnt] = Edge(u, v, w, g[u]), g[u] = ecnt
#define MID (l + r >> 1)
#define mst(a, v) memset(a, v, sizeof(a))
#define bg(x)            
	Edge edg[maxn << x]; 
	int g[maxn], ecnt
#define ex(v)  
	cout << v; 
	return 0
#define debug(x) cout << "debug: " << x << endl;
#define sqr(x) ((x) * (x))

using namespace std;
typedef long long ll;
typedef unsigned long long ull;
typedef double db;
typedef long double ld;
typedef pair<int, int> pii;
typedef pair<ll, ll> pll;
typedef complex<double> cpx;
typedef vector<int> vi;
typedef vector<ll> vll;
typedef map<int, int> mii;
typedef map<ll, ll> mll;

char READ_DATA;
int SIGNAL_INPUT;
template <typename Type>
inline Type ru(Type &v)
{
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
		v = v * 10 + READ_DATA - '0';
	v *= SIGNAL_INPUT;
	return v;
}
inline ll modru(ll &v)
{
	ll p = 0;
	SIGNAL_INPUT = 1;
	while ((READ_DATA = getchar()) < '0' || READ_DATA > '9')
		if (READ_DATA == '-')
			SIGNAL_INPUT = -1;
		else if (READ_DATA == EOF)
			return EOF;
	p = v = READ_DATA - '0';
	while ((READ_DATA = getchar()) >= '0' && READ_DATA <= '9')
	{
		v = (v * 10 + READ_DATA - '0') % mod;
		p = (p * 10 + READ_DATA - '0') % (mod - 1);
	}
	v *= SIGNAL_INPUT;
	return p;
}
template <typename A, typename B>
inline int ru(A &x, B &y)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	return 2;
}
template <typename A, typename B, typename C>
inline int ru(A &x, B &y, C &z)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	return 3;
}
template <typename A, typename B, typename C, typename D>
inline int ru(A &x, B &y, C &z, D &w)
{
	if (ru(x) == EOF)
		return EOF;
	ru(y);
	ru(z);
	ru(w);
	return 4;
}
inline ll gcd(ll a, ll b)
{
	while (b)
	{
		a %= b;
		swap(a, b);
	}
	return a;
}

inline ll fastmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return (a * b - (ll)((long double)a * b / mod) * mod + mod) % mod;
}

inline ll dirmul(ll a, ll b, ll mod = 1e9 + 7)
{
	return a * b % mod;
}

inline ll ss(ll a, ll b, ll mod = 1e9 + 7, ll(*mul)(ll, ll, ll) = dirmul)
{
	if (b < 0)
	{
		b = -b;
		a = ss(a, mod - 2, mod);
	}
	ll ans = 1;
	while (b)
	{
		if (b & 1)
			ans = mul(ans, a, mod);
		a = mul(a, a, mod);
		b >>= 1;
	}
	return ans;
}
inline int isprime(ll n)
{
	if (n == 1)
		return 0;

	for (ll d = 2; d * d <= n; ++d)
	{
		if (n % d == 0)
			return 0;
	}

	return 1;
}

template <typename Type>
void brc(Type *a, int n)
{
	int k;
	for (int i = 1, j = n / 2; i < n - 1; i++)
	{
		if (i < j)
			swap(a[i], a[j]);

		k = n >> 1;
		while (j >= k)
		{
			j ^= k;
			k >>= 1;
		}
		if (j < k)
			j ^= k;
	}
}
void fft(cpx *a, int n, int inv = 1)
{
	cpx u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		cpx wn(cos(inv * 2.0 * PI / h), sin(inv * 2.0 * PI / h));
		for (int j = 0; j < n; j += h)
		{
			cpx w(1, 0);
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)];
				a[k] = u + t;
				a[k + (h >> 1)] = u - t;
				w *= wn;
			}
		}
	}
	if (inv == -1)
		RP(i, 0, n)
		a[i] /= n;
}
void ntt(ll *a, int n, int inv = 1)
{
	ll u, t;
	brc(a, n);
	for (int h = 2; h <= n; h <<= 1)
	{
		ll wn = ss(3, (mod - 1) / h);
		if (inv == -1)
			wn = ss(wn, mod - 2);
		for (int j = 0; j < n; j += h)
		{
			ll w = 1;
			for (int k = j; k < j + (h >> 1); k++)
			{
				u = a[k];
				t = w * a[k + (h >> 1)] % mod;
				a[k] = (u + t) % mod;
				a[k + (h >> 1)] = (u - t + mod) % mod;
				(w *= wn) %= mod;
			}
		}
	}
	if (inv == -1)
	{
		ll tmp = ss(n, mod - 2);
		RP(i, 0, n)
			(a[i] *= tmp) %= mod;
	}
}
struct Edge
{
	int u, v, nxt;
	//ll w;
	Edge(int _u = 0, int _v = 0, /*ll _w = 0,*/ int _nxt = 0)
	{
		u = _u;
		v = _v;
		//w = _w;
		nxt = _nxt;
	}

	/*int operator<(const Edge &b) const
	{
		return w < b.w;
	}*/
};
struct CMP
{
	int operator()(const int &a, const int &b) const
	{
		return a > b;
	}
};

const int maxn = 1e3 + 5;
/*------------------------------------------------------------------------yah01------------------------------------------------------------------------*/

int n, m;
int f[51][51], g[51][51][51];
int main()
{
	int T;
	ru(T);
	rp(t, 1, T)
	{
		mst(f, 0);
		mst(g, 0);
		ru(n, m);
		while (m--)
		{
			int u, v;
			ru(u, v);
			f[u][v] = f[v][u] = 1;
		}

		ll ans = 0, tot = ss(2, n);
		tot -= (1 + n + n * (n - 1)*ss(2,mod-2)) % mod;
		if (tot < 0) tot += mod;

		rp(i, 1, n)
		{
			rp(j, i + 1, n)
			{
				rp(k, j + 1, n)
				{
					if ((f[i][j] && f[i][k] && f[j][k]) || (!f[i][j] && !f[i][k] && !f[j][k]))
					{
						++ans;
						g[i][j][k] = 1;
					}
					--tot;
					if (tot < 0) tot += mod;
				}
			}
		}

		rp(i, 1, n)
		{
			rp(j, i + 1, n)
			{
				rp(k, j + 1, n)
				{
					rp(p, k + 1, n)
					{
						if (g[i][j][k] || g[i][j][p] || g[j][k][p] || g[i][k][p])
						{
							++ans;
							if (ans >= mod) ans -= mod;
						}
						--tot;
						if (tot < 0) tot += mod;
					}
				}
			}
		}

		rp(i, 1, n)
		{
			rp(j, i + 1, n)
			{
				rp(k, j + 1, n)
				{
					rp(p, k + 1, n)
					{
						rp(q, p + 1, n)
						{
							if (g[i][j][k] || g[i][j][p] || g[i][j][q] || g[i][k][p]||g[i][k][q] || g[i][p][q] || g[j][k][p] || g[j][k][q] || g[j][p][q] || g[k][p][q])
							{
								++ans;
								if (ans >= mod) ans -= mod;
							}
							--tot;
							if (tot < 0) tot += mod;
						}
					}
				}
			}
		}

		ans += tot;
		ans %= mod;
		printf("Case #%d: %lld
", t, ans);
	}
	return 0;
}

/*
5
2 4 6 8 10
-1 2 2 2
1000
+ 1 10
+ 1 -2
+ 1 1
*/

    E - The Fastest Runner Ms. Zhang (jmx赛后补题)

    题目来源：HDU-5917

    赛后补题

    显然这是一个环套树的问题。

    定义$dis_{u,v}$为$u$和$v$的距离。定义$len$为环的长度。定义$mxdep_u$为以环上的点$u$为根的树的最深深度。以$2n$为初始标准，对于我们选定的点对$(S,T)$，有如下两种情况:

    1. 如果$(S,T)$在同一颗子树中，那么答案就是$2n-len-dis_{S,T}$
    2. 如果$(S,T)$不在同一颗子树中，令$S$和$T$所在的树的根分别为$U$和$V$，那么答案就是$2n-mxdep_S-mxdep_T-dis_{U,V}(取环上较长的路径)-2$

    从两种方法中取最大值就行了。前一类是经典问题，后一类可以将mxdep全部求出，然后将环复制两份变成链，用单调队列或者线段树处理（忘了单调队列怎么写所以写的线段树）。不过输出方案很麻烦。

    #include <bits/stdc++.h>
#define ll long long
#define inf 0x6fffffff
#define N 200086
using namespace std;
int read(){
    int x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
int n,ans,ansx,ansy;
int inl[N],vis[N],head[N],dep[N],mxdep[N],v[N];
int q[N*2],nxt[N*2];
int p[N];
int tt;
bool flag;
int len;
int l[N];
int f[N*4];
void clr()
{
    ans=tt=flag=len=0;
    for(int i=1;i<=n;i++)head[i]=vis[i]=inl[i]=dep[i]=mxdep[i]=0;
}
void ins(int u,int v){q[++tt]=v;nxt[tt]=head[u];head[u]=tt;}
void findcircle(int i,int pre)
{
    vis[i]=1;p[i]=pre;
    for(int j=head[i];j;j=nxt[j])
        if(q[j]!=pre)
        {
            if(vis[q[j]])
            {
                for(int x=i;x!=q[j];x=p[x]){inl[x]=1;l[++len]=x;}
                l[++len]=q[j];inl[q[j]]=1;
                flag=1;return;
            }
            else findcircle(q[j],i);
            if(flag)return;
        }
}
void update(int tmp,int x,int y)
{
    if(x>y)swap(x,y);
    if(tmp<ans)return;
    if(tmp>ans)
    {
        ans=tmp;ansx=x;ansy=y;
    }
    else
    {
        if(x>ansx)return;
        if(x<ansx)
        {
            ansx=x;ansy=y;
        }
        else{
            if(y<ansy)ansy=y;
        }
    }
}
void dfs(int i,int ro,int pre)
{
    mxdep[i]=i;
    int id=i;
    for(int j=head[i];j;j=nxt[j])
        if(!inl[q[j]]&&q[j]!=pre)
        {
            dep[q[j]]=dep[i]+1;dfs(q[j],ro,i);
            int tmp=dep[id]+dep[mxdep[q[j]]]-2*dep[i];
            update(tmp+len,mxdep[q[j]],id);
            if(dep[mxdep[q[j]]]>dep[id]||(dep[mxdep[q[j]]]==dep[id]&&mxdep[q[j]]<id))id=mxdep[q[j]];
        }
    mxdep[i]=id;
}
int findmn(int a,int b)
{
    if(a==0||b==0)return a+b;
    if(v[a]<v[b])return a;
    if(v[a]>v[b])return b;
    if(mxdep[l[a]]<mxdep[l[b]])return a;else return b;
}
void build(int i,int l,int r)
{
    if(l==r){f[i]=l;return;}
    int mid=(l+r)/2;
    build(i*2,l,mid);build(i*2+1,mid+1,r);
    f[i]=findmn(f[i*2],f[i*2+1]);
}
int ask(int i,int l,int r,int lt,int rt)
{
    if(lt<=l&&r<=rt)return f[i];
    int mid=(l+r)/2,res=0;
    if(lt<=mid)res=findmn(res,ask(i*2,l,mid,lt,rt));
    if(mid+1<=rt)res=findmn(res,ask(i*2+1,mid+1,r,lt,rt));
    return res;
}
int main()
{
    int T=read();
    for(int cas=1;cas<=T;cas++)
    {
        n=read();
        clr();
        for(int i=1;i<=n;i++)
        {
            int u=read(),v=read();
            ins(u,v);ins(v,u);
        }
        findcircle(1,0);
        for(int i=1;i<=n;i++)
            if(inl[i]){mxdep[i]=i;dfs(i,i,0);}
        //for(int i=1;i<=len;i++)printf("%d %d
",l[i],mxdep[l[i]]);
        for(int i=1;i<=len;i++)l[i+len]=l[i];
        len*=2;
        for(int i=1;i<=len;i++)v[i]=i-dep[mxdep[l[i]]];
        build(1,1,len);
        for(int i=2;i<=len;i++)
        {
            int tmp=i+dep[mxdep[l[i]]]+2;
            int id=ask(1,1,len,max(1,i-len/2+1),i-1);
            update(tmp-v[id],mxdep[l[id]],mxdep[l[i]]);
        }
        printf("Case #%d: %d %d %d
",cas,2*n-ans,ansx,ansy);
    }
    return 0;
}

    K - Binary Indexed Tree (jmx赛后补题)

    题目来源：HDU-5921

    赛后补题

    对于每一个数计算贡献，即这个数被really change了多少次。显然，不以某个数为前缀的数对个数即为这个点的贡献。对于一个数$x$，如果在$n$以内的以$x$为前缀的数的个数为$t$，那么对答案的贡献就是$t imes (n-t+1)$。

    对于大多数数字，$t$的取值都是$2j$。可以使用数位dp来统计这种情况。$f_{i,j,k}$表示$i$位二进制数结尾有恰好$j$个$0$并且最高位为$k$的二进制数个数。这样的话用一个普通数位dp就能统计出来$t$取值为$2j$的数的个数，然后统一算贡献。


    那些$t$的取值不是$2$的次幂的数，一定是$n$的前缀。对于$n$的前缀$x$，$t=n-x+1$。把这种情况再算一下加起来就好。

    #include <bits/stdc++.h>
#define MOD 1000000007
#define inf 0x6fffffff
using namespace std;
typedef long long ll;
ll read(){
    ll x=0,f=1;char ch=getchar();
    while(ch<'0'||ch>'9'){if(ch=='-')f=-1;ch=getchar();}
    while(ch>='0'&&ch<='9'){x=x*10+ch-'0';ch=getchar();}
    return x*f;
}
ll f[61][61][2],bin[61];
void pre()
{
    bin[0]=1;
    for(int i=1;i<=60;i++)bin[i]=(bin[i-1]<<1)%MOD;
    for(int i=1;i<=60;i++)
    {
        for(int j=0;j<=i-2;j++)
        {
            f[i][j][0]=bin[i-j-2];
            f[i][j][1]=bin[i-j-2];
        }
        f[i][i-1][1]=1;
        f[i][i][0]=1;
    }
}
ll calc(ll n)
{
    int num[61],cnt=0;
    ll x=n;
    while(x)
    {
        num[++cnt]=x&1;x>>=1;
    }
    ll ans=0;
    n%=MOD;
    for(int k=0;k<=cnt;k++)
    {
        ll res=0;
        for(int i=1;i<cnt;i++)res=(res+f[i][k][1])%MOD;
        for(int i=cnt-1;i;i--)
        {
            if(num[i]==1)
            {
                if(k<i-1)res=(res+f[i][k][0])%MOD;
            }
        }
        res=res*(bin[k]*((n+1-bin[k])%MOD)%MOD)%MOD;
        ans=(ans+res)%MOD;
    }
    ll tmp=(1LL<<(cnt-1))%MOD;
    for(int i=cnt-1;i;i--)
    {
        if(num[i+1])ans=(ans+(n-tmp+1)*(n+1-(n-tmp+1))%MOD)%MOD;
        tmp=(tmp+num[i]*(1LL<<(i-1))%MOD)%MOD;
    }
    if(num[1])ans+=n;
    return (ans+MOD)%MOD;
}
int main()
{
    pre();
    int T=read();
    for(int cas=1;cas<=T;cas++)
    {
        ll n=read();
        printf("Case #%d: ",cas);
        cout<<calc(n)<<endl;
    }
    return 0;
}
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  • 原文地址：https://www.cnblogs.com/oldjang/p/10657796.html
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