题解报告（CDUT暑期集训——第一场）

题解报告（CDUT暑期集训——第一场）

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A - Maximum Multiple

HDU - 6298

• 思路：先按照题意打表 发现规律 就出来了（最开始没开long long贡献了3发 然后又忘了换行又贡献了一发

• AC代码

---

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll Pow(ll a, ll b, ll c){
    ll ans = 1;
    a %= c;
    while (b){
        if (b & 1) ans = (ans * a) % c;
        a = (a * a) % c;
        b >>= 1;
    }
    return (ans % c);
}

bool cmp(const int &a, const int &b){
    return a < b;
}

ll gcd(ll x, ll y){
    return x % y == 0 ? y : gcd(y, x % y);
}

ll lcm(ll x, ll y){
    return x / gcd(x, y) * y;
}

const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];

void init(){
	for (int i = 0; i < MAXN; i ++ ){
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; j ++ ){
			comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
			comb[i][j] %= mod;
		}
	}
}

ll mult_mod(ll a, ll b, ll c){
    a %= c;
    b %= c;
    ll res = 0;
    while (b){
        if (b & 1){
            res += a;
            res %= c;
        }
        a <<= 1;
        a %= c;
        b >>= 1;
    }
    return res;
}

int main(){
    ll t;
    scanf("%lld", &t);
    ll n;
    while (t--){
        scanf("%lld", &n);
        if (n % 12 == 1 || n % 12 == 2 || n % 12 == 5 || n % 12 == 7 || n % 12 == 10 || n % 12 == 11) printf("-1
");
        else{
            if (n % 3) printf("%lld", (n / 4) * (n / 4) * (n / 4) * 2);
            else printf("%lld", (n / 3) * (n / 3) * (n / 3));
        }
    }
    return 0;
}

B - Balanced Sequence

HDU - 6299

• 思路：贪心 先把每个串里面的括号匹配之后去掉 再对它们排序 匹配剩下的括号就行了（开始莫名其妙MLE了九发 找不到原因 然后再交TLE RE 后面发现cmp写错了 改了之后就过了

• AC代码

---

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string>
#include<string.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

const int N = 100010;

int n, t, ans, cnt_l;
char c[N];

struct st{
    int l;
    int r;
}s[N];

bool cmp(st a, st b){
    if (a.l > a.r && b.l <= b.r) return true;
    else if (a.l <= a.r && b.l > b.r) return false;
    else if (a.l > a.r && b.l > b.r) return a.r < b.r;
    else return a.l > b.l;
}

int main(){
    scanf("%d", &t);
    while (t --){
        ans = 0;
        cnt_l = 0;
        scanf("%d", &n);
        for (int i = 1; i <= n; i ++ ){
            s[i].l = 0;
            s[i].r = 0;
            scanf("%s", c);
            for (int j = 0; j < strlen(c); j ++ ){
                if (c[j] == '(') s[i].l ++;
                if (c[j] == ')') s[i].l == 0 ? s[i].r ++ : (s[i].l -= 1, ans += 2);
            }
        }
        sort(s + 1, s + n + 1, cmp);
        for (int i = 1; i <= n; i ++ ){
            if (s[i].r > cnt_l) s[i].r = cnt_l;
            ans += 2 * s[i].r;
            cnt_l += s[i].l - s[i].r;
        }
        printf("%d
", ans);
    }
    return 0;
}

C - Triangle Partition

HDU - 6300

• 思路：贪心 写个struct对x排个序就出来了（

• AC代码

---

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll Pow(ll a, ll b, ll c){
    ll ans = 1;
    a %= c;
    while (b){
        if (b & 1) ans = (ans * a) % c;
        a = (a * a) % c;
        b >>= 1;
    }
    return (ans % c);
}

//bool cmp(const int &a, const int &b){
//    return a < b;
//}

ll gcd(ll x, ll y){
    return x % y == 0 ? y : gcd(y, x % y);
}

ll lcm(ll x, ll y){
    return x / gcd(x, y) * y;
}

const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];

void init(){
	for (int i = 0; i < MAXN; i ++ ){
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; j ++ ){
			comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
			comb[i][j] %= mod;
		}
	}
}

ll mult_mod(ll a, ll b, ll c){
    a %= c;
    b %= c;
    ll res = 0;
    while (b){
        if (b & 1){
            res += a;
            res %= c;
        }
        a <<= 1;
        a %= c;
        b >>= 1;
    }
    return res;
}

const int N = 1010;

struct point{
    int x, y;
    int index;
}p[N];

bool cmp(point a, point b){
    if (a.x == b.x) return a.y < b.y;
    return a.x < b.x;
}

int main(){
    int t;
    scanf("%d", &t);
    int n;
    while (t--){
        scanf("%d", &n);
        for (int i = 1; i <= 3 * n; i ++ ){
            scanf("%d%d", &p[i].x, &p[i].y);
            p[i].index = i;
        }
        sort(p + 1, p + 3 * n + 1, cmp);
        for (int i = 1; i <= n; i ++ )
            printf("%d %d %d
", p[(i - 1) * 3 + 1].index, p[(i - 1) * 3 + 2].index, p[(i - 1) * 3 + 3].index);
    }
    return 0;
}

D - Distinct Values

HDU - 6301

• __思路：贪心 先排序 后用set维护就行了（学到了一个新东西 迭代器赋值前要加* 要不然会报错 __

• AC代码

---

#include<stdio.h>
#include<iostream>
#include<math.h>
#include<algorithm>
#include<string.h>
#include<queue>
#include<set>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll Pow_mod(ll a, ll b, ll c){
    ll ans = 1;
    a %= c;
    while (b){
        if (b & 1) ans = (ans * a) % c;
        a = (a * a) % c;
        b >>= 1;
    }
    return (ans % c);
}

const int N = 1000010;

int t, n, m, L, R;
set<int> st;
int ans[N];

struct node{
    int l, r;
}interval[N];

int cmp(node a, node b){
    return a.l == b.l ? a.r < b.r : a.l < b.l;
}

int main(){
    scanf("%d", &t);
    while (t -- ){
        st.clear();
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= m; i ++ )
            scanf("%d%d", &interval[i].l, &interval[i].r);
        sort(interval + 1, interval + m + 1, cmp);
        L = interval[1].l;
        R = interval[1].r;
        for (int i = 1; i <= n; i ++ ){
            st.insert(i);
            ans[i] = 1;
        }
        for (int i = L; i <= R; i ++ ){
            ans[i] = *st.begin();
            st.erase(st.begin());
        }
        for (int i = 2; i <= m; i ++ ){
            while (L < interval[i].l){
                st.insert(ans[L]);
                L ++;
            }
            while (R < interval[i].r){
                if (R < interval[i].l - 1) R ++;
                else{
                    R ++;
                    ans[R] = *st.begin();
                    st.erase(st.begin());
                }
            }
        }
        for (int i = 1; i < n; i ++ ) printf("%d ", ans[i]);
        printf("%d
", ans[n]);
    }
}

G - Chiaki Sequence Revisited

HDU - 6304

• 思路：先打表 找规律 令f(i)为i出现的次数 再打表 去掉a1 = 1出现的一次 可以得到f(2 * i) = f(i) + 1 f(2 * i + 1) = 1 当时到了这里就卡死了 卡了两个半小时 然后想着把f(i) = n, n = 1, 2, 3, ...打印出来看下 结果一打印就发现了f(i)相同的构成了一个首项为2^(n-1) 公差为2^n的等差数列 就直接根据公式就解出来了 结束前没能写出来 结束之后几分钟写好交了三发才过 第一发用自己写的乘法取膜函数tle了 第二发有一处忘了取膜贡献了一发wa 第三发才终于过了

• AC代码

---

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll Pow_mod(ll a, ll b, ll c){
    ll ans = 1;
    a %= c;
    while (b){
        if (b & 1) ans = (ans * a) % c;
        a = (a * a) % c;
        b >>= 1;
    }
    return (ans % c);
}

ll Pow(ll a, ll b){
    ll ans = 1;
    while (b){
        if (b & 1) ans *= a;
        a *= a;
        b >>= 1;
    }
    return ans;
}

bool cmp(const int &a, const int &b){
    return a < b;
}

ll gcd(ll x, ll y){
    return x % y == 0 ? y : gcd(y, x % y);
}

ll lcm(ll x, ll y){
    return x / gcd(x, y) * y;
}

const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];

void init(){
	for (int i = 0; i < MAXN; i ++ ){
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; j ++ ){
			comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
			comb[i][j] %= mod;
		}
	}
}

ll mult_mod(ll a, ll b, ll c){
    a %= c;
    b %= c;
    ll res = 0;
    while (b){
        if (b & 1){
            res += a;
            res %= c;
        }
        a <<= 1;
        a %= c;
        b >>= 1;
    }
    return res;
}

const int N = 63;

ll s[N], t[N];

ll calc(ll n){
    ll tmp = 0;
    if (n == 1) return 1;
    n -= 1;
    for (int i = N - 1; i >= 0; i -- ){
        while (s[i] <= n){
            n -= s[i];
            tmp += t[i];
        }
    }
    return tmp;
}

ll get_pos(ll n){
    ll pos = 0;
    if (n == 1) return 1;
    n -= 1;
    for (int i = N - 1; i >= 0; i -- ){
        while (t[i] <= n){
            n -= t[i];
            pos += s[i];
        }
    }
    return pos + 1;
}

int main(){
    s[0] = 1;
    t[0] = 1;
    for (int i = 1; i < N; i++){
        s[i] = s[i - 1] * 2 + 1;
        t[i] = t[i - 1] * 2;
    }
    int T;
    scanf("%d", &T);
    ll n;
    while (T--){
        scanf("%lld", &n);
        ll tmp = calc(n);
        ll cnt = n - get_pos(tmp);
        ll sum = 0;
        for (int i = 1; t[i - 1] <= tmp; i ++ ){
            ll b1 = t[i - 1];
            ll d = t[i];
            ll n_ = (tmp - b1) / d;
            if ((tmp - b1) % d) n_ += 1;
            ll bn = b1 + (n_ - 1) * d;
            ll sum_b = (b1 % mod + bn % mod) * (n_ % mod) * 500000004 % mod;
            sum += i * sum_b % mod;
            sum %= mod;
            if (!((tmp - b1) % d)) sum += (cnt % mod) * (tmp % mod) % mod;
            sum %= mod;
        }
        printf("%lld
", sum + 1);
    }
    return 0;
}

K - Time Zone

HDU - 6308

• 思路：数学题 把时区差换算成时间差（分钟） 然后换算过去就行了 最开始写了个贼蠢的模拟 写的很长 还wa了 后面才想起来没写%02d才wa的

• AC代码

---

#include<bits/stdc++.h>
typedef long long ll;
typedef unsigned long long ull;
using namespace std;

ll Pow(ll a, ll b, ll c){
    ll ans = 1;
    a %= c;
    while (b){
        if (b & 1) ans = (ans * a) % c;
        a = (a * a) % c;
        b >>= 1;
    }
    return (ans % c);
}

bool cmp(const int &a, const int &b){
    return a < b;
}

ll gcd(ll x, ll y){
    return x % y == 0 ? y : gcd(y, x % y);
}

ll lcm(ll x, ll y){
    return x / gcd(x, y) * y;
}

const int MAXN = 2000 + 5;
const int mod = 1e9 + 7;
ll comb[MAXN][MAXN];

void init(){
	for (int i = 0; i < MAXN; i ++ ){
		comb[i][0] = comb[i][i] = 1;
		for (int j = 1; j < i; j ++ ){
			comb[i][j] = comb[i - 1][j] + comb[i - 1][j - 1];
			comb[i][j] %= mod;
		}
	}
}

ll mult_mod(ll a, ll b, ll c){
    a %= c;
    b %= c;
    ll res = 0;
    while (b){
        if (b & 1){
            res += a;
            res %= c;
        }
        a <<= 1;
        a %= c;
        b >>= 1;
    }
    return res;
}

int main(){
    int t;
    scanf("%d", &t);
    int a, b;
    while (t--){
        char flag;
        double utc;
        scanf("%d %d UTC%c%lf", &a, &b, &flag, &utc);
        int tmp = (int)(utc * 10 + 0.1);
        if (flag == '-') tmp = -tmp;
        int sum = a * 60 + b + (tmp * 6 - 8 * 60);
        sum %= (24 * 60);
        if (sum < 0) sum += (24 * 60);
        printf("%02d:%02d
", sum / 60, sum % 60);
    }
    return 0;
}