2021百度之星复赛

前400名有衣服拿，然后我
草

原本没想过能拿衣服 但现在差点拿衣服有点难受

签到题漏特判wa*2 过完签到jyz表示他在外边 用txt敲了代码让我帮忙编译 时间--

第二题漏情况了样例过不去 调了半天hhhhh

1002

枚举k表示用乘号将n个加号划分为k个集合，然后在求一下插入乘号的方案即可，也是用斯特林数
群友表示这已经是csp初中组的知识点了

#include<bits/stdc++.h>
using namespace std;
const int maxn = 3e3 + 7, md = 1e9 + 7;
#define ll long long
int rd() {
    int s = 0, f = 1; char c = getchar();
    while (c < '0' || c > '9') {if (c == '-') f = -1; c = getchar();}
    while (c >= '0' && c <= '9') {s = s * 10 + c - '0'; c = getchar();}
    return s * f;
}
int n, m, k, tot, a[maxn], T;
ll S[maxn][maxn], fac[maxn];
ll ans = 0;
int main() {
    T = rd();
    S[0][0] = 1;
    fac[0] = fac[1] = 1;
    for (int i = 1; i <= 3000; i++) {
        fac[i] = fac[i-1] * i % md;
        for (int j = 1; j <= 3000; j++) {
            S[i][j] = (S[i-1][j-1] + 1ll*j*S[i-1][j]%md)%md;
        }
    }
    while (T--) {
        ans = 0;
        n = rd(), m = rd();
        for (k = 1; k <= min(n, m+1); k++) {
            ll res = (S[m][k-1] * fac[k-1])%md;
            if (k<=m) res = (res + 2ll * fac[k]%md * S[m][k] % md);
            if (k+1 <= m) res = (res + fac[k+1] * S[m][k+1] % md) % md;
            ans = (ans + fac[k] * S[n][k] % md * res % md) % md;
        }
        printf("%lld\n", ans);
    }
}

标签：题漏,int,long,签到,while,2021,之星,复赛,getchar
来源： https://www.cnblogs.com/YjmStr/p/15171092.html