【AtCoder】AGC032

AGC032

A - Limited Insertion

这题就是从后面找一个最靠后而且当前可以放的，可以放的条件是它的前面正好放了它的数值-1个数

如果不符合条件就退出

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
int a[MAXN],N,cnt[MAXN];
bool vis[MAXN];
void Solve() {
    read(N);
    for(int i = 1 ; i <= N ; ++i) read(a[i]);
    for(int i = 1 ; i <= N ; ++i) {
        int cnt = 0;
        for(int j = 1 ; j < i ; ++j) {
            if(a[j] <= a[i]) ++cnt;
        }
        if(cnt + 1 < a[i]) {puts("-1");return;}
    }
    for(int i = 1 ; i <= N ; ++i) {
        cnt[0] = 0;
        for(int j = 1 ; j <= N ; ++j) cnt[j] = cnt[j - 1] + vis[j];
        for(int j = N ; j >= 1 ; --j) {
            if(!vis[j] && cnt[j - 1] + 1 == a[j]) {
                vis[j] = 1;
                out(a[j]);enter;
                break;
            }
        }
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

B - Balanced Neighbors

如果是偶数个，分成和相等的(frac{N}{2})份

如果是奇数个，最后一个点单独为一组，前(N - 1)个两个一组分成和相等的(frac{N - 1}{2})份

然后把一组作为一个点，建一个完全图，两组(一组1,2，一组3,4)之间连边就是

1-2,1-4,2-3,2-4

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
vector<pii > v;
void Solve() {
    read(N);
    if(N & 1) {
        for(int i = 1 ; i < N ; ++i) {
            v.pb(mp(i,N));
        }
        --N;
    }
    for(int i = 1 ; i <= N ; ++i) {
        for(int j = i + 1 ; j <= N ; ++j) {
            if(i + j != N + 1) v.pb(mp(i,j));
        }
    }
    out(v.size());enter;
    for(auto t : v) {
        out(t.fi);space;out(t.se);enter;
    }
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

C - Three Circuits

必定存在一个欧拉回路

如果一个点有六个或以上点度必定存在

如果有三个点以上有四个点度必定存在

如果只有两个点有四个点度

那么如果这四条路都在这两个点之间，那么就无解

否则有解

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 100005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
struct node {
    int to,next;
}E[MAXN * 2];
int head[MAXN],sumE;
int N,M;
int cnt[MAXN];
bool vis[MAXN];
void add(int u,int v) {
    E[++sumE].to = v;
    E[sumE].next = head[u];
    head[u] = sumE;
}
void dfs(int u) {
    vis[u] = 1;
    for(int i = head[u] ; i ; i = E[i].next) {
        int v = E[i].to;
        if(!vis[v]) dfs(v);
    }
}
void Solve() {
    read(N);read(M);
    int a,b;
    for(int i = 1 ; i <= M ; ++i) {
        read(a);read(b);
        cnt[a]++;cnt[b]++;
        add(a,b);add(b,a);
    }
    for(int i = 1 ; i <= N ; ++i) {
        if(cnt[i] & 1) {puts("No");return;}
    }
    int t = 0;
    int p = 0,q = 0;
    for(int i = 1 ; i <= N ; ++i) {
        if(cnt[i] >= 6) {puts("Yes");return;}
        if(cnt[i] >= 4) {
            ++t;
            if(!p) p = i;
            else if(!q) q = i;
        }
    }
    if(t > 2) {puts("Yes");return;}
    if(t == 2) {
        vis[p] = 1;
        dfs(q);
        for(int i = 1 ; i <= N ; ++i) {
            if(!vis[i]) {puts("Yes");return;}
        }
    }
    puts("No");
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

D - Rotation Sort

把操作改成数轴上，我可以把一个点移动到左边或右边任意一个位置上，可以不必要是整数点

显然对于每个点操作只进行一次

然后拆成((-infty,1),(1,2),(2,3),(3,4)....(N - 1,N),(N,+infty))和整数点(1,2,3,4,5,6..N)

然后根据位置dp即可

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 5005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
int N;
int64 A,B;
int64 dp[MAXN][2 * MAXN],s[MAXN][2 * MAXN];
int p[MAXN],pos[MAXN];
void Solve() {
    read(N);read(A);read(B);
    for(int i = 1 ; i <= N ; ++i) {read(p[i]);pos[p[i]] = i;}
    for(int i = 1 ; i <= N ; ++i) {
        s[i][0] = 1e18;
        for(int j = 1 ; j <= 2 * N + 1 ; ++j) {
            if(j & 1) {
                dp[i][j] = s[i - 1][j] + (j < pos[i] * 2 ? B : A);
            }
            else {
                dp[i][j] = s[i - 1][j - 1] + (j != pos[i] * 2 ? (j < pos[i] * 2 ? B : A): 0);
            }
            s[i][j] = min(s[i][j - 1],dp[i][j]);
        }
    }
    out(s[N][2 * N + 1]);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

E - Modulo Pairing

用蓝色表示相加小于M的，红色表示大于M的

然后就会变成前面是蓝的，后面是红的

简单分析发现蓝的越小越好，红的越大越好，所以求出能向左最长的红色序列

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 200005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
        if(c == '-') f = -1;
        c = getchar();
    }
    while(c >= '0' && c <= '9') {
        res = res * 10 +c - '0';
        c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
        out(x / 10);
    }
    putchar('0' + x % 10);
}
int N,M;
int a[MAXN];
void Solve() {
    read(N);read(M);
    for(int i = 1 ; i <= 2 * N ; ++i) read(a[i]);
    sort(a + 1,a + 2 * N + 1);
    int r = -1;
    for(int i = 2 * N ; i >= 1 ; --i) {
        int t = lower_bound(a + 1,a + 2 * N + 1,M - a[i]) - a;
        if((t & 1) == (i & 1)) ++t;
        if(r == -1) r = i + t;
        else r = max(r,i + t);
    }
    if(r == -1) r = 4 * N + 1;
    int ans = 0;
    for(int i = 2 * N ; i >= 1 ; --i) {
        if(r - i < i) ans = max(ans,(a[i] + a[r - i]) % M);
        else break;
    }
    r = 1 + r - 2 * N - 1;
    for(int i = 1 ; i <= 2 * N ; ++i) {
        if(i < r - i) ans = max(ans,a[i] + a[r - i]);
        else break;
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
    return 0;
}

F - One Third

切一刀，画一条红线，然后正120画一条蓝线，反120画一条绿线

我们要求的就是任意1/3中最短两个颜色不同之间的线的角度大小

然后我们取第一刀红线和蓝线之间的 1/3区间，然后里面被分成了(N)份，我们递推一个(f(i))，表示有(i)份左右两边区间颜色不同的概率

然后(g(i))表示(i)份中最短的一段的期望长度，(frac{1}{3})长度里(N)份中选(i)份期望长度是(frac{i}{3N})，可以认为分成(N)份中选(i)份每个点被选中的概率是(frac{i}{N} = frac{inom{i - 1}{N - 1}}{inom{i}{N}})

那(i)段中的期望最小长度呢

是(E(min) = int_{t = 0}^{1/i}P(min geq t) dt = int_{t = 0}^{1/i}(1 - it)^{i - 1}dt = int_{t = 0}^{1}frac{t^{i - 1}}{i} dt = frac{1}{i^2})

所以(g(i) = frac{1}{3iN})

答案就是所有的(f(i)g(i))的和

#include <bits/stdc++.h>
#define fi first
#define se second
#define pii pair<int,int>
#define mp make_pair
#define pb push_back
#define space putchar(' ')
#define enter putchar('
')
#define eps 1e-10
#define MAXN 1000005
//#define ivorysi
using namespace std;
typedef long long int64;
typedef unsigned int u32;
typedef double db;
template<class T>
void read(T &res) {
    res = 0;T f = 1;char c = getchar();
    while(c < '0' || c > '9') {
	if(c == '-') f = -1;
	c = getchar();
    }
    while(c >= '0' && c <= '9') {
	res = res * 10 +c - '0';
	c = getchar();
    }
    res *= f;
}
template<class T>
void out(T x) {
    if(x < 0) {x = -x;putchar('-');}
    if(x >= 10) {
	out(x / 10);
    }
    putchar('0' + x % 10);
}
const int MOD = 1000000007;
int N;
int f[MAXN][3],g[MAXN],fac[MAXN],invfac[MAXN],inv[MAXN];
int inc(int a,int b) {
    return a + b >= MOD ? a + b - MOD : a + b;
}
int mul(int a,int b) {
    return 1LL * a  * b % MOD;
}
int fpow(int x,int c) {
    int res = 1,t = x;
    while(c) {
	if(c & 1) res = mul(res,t);
	t = mul(t,t);
	c >>= 1;
    }
    return res;
}
int C(int n,int m) {
    if(n < m) return 0;
    return mul(fac[n],mul(invfac[m],invfac[n - m]));
}
void Solve() {
    read(N);
    fac[0] = 1;
    for(int i = 1 ; i <= N; ++i) fac[i] = mul(fac[i - 1],i);
    invfac[N] = fpow(fac[N],MOD - 2);
    for(int i = N - 1 ; i >= 0 ; --i) invfac[i] = mul(invfac[i + 1],i + 1);
    inv[1] = 1;
    for(int i = 2 ; i <= 1000000 ; ++i) inv[i] = mul(inv[MOD % i],MOD - MOD / i);
    f[0][0] = 1;
    int iv3= fpow(inv[3],N - 1);
    int ans = 0;
    for(int i = 1 ; i <= N ; ++i) {
	f[i][1] = inc(f[i - 1][0],f[i - 1][2]);
	f[i][0] = inc(f[i - 1][1],f[i - 1][2]);
	f[i][2] = inc(f[i - 1][0],f[i - 1][1]);
	g[i] = mul(mul(inv[3],inv[N]),inv[i]);
	int p = mul(mul(f[i][1],C(N,i)),iv3);
	ans = inc(ans,mul(g[i],p));
    }
    out(ans);enter;
}
int main() {
#ifdef ivorysi
    freopen("f1.in","r",stdin);
#endif
    Solve();
}