hdu 5894 (组合)

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得到公式为：n * C(n - 1- m * k,m - 1) * n / m，记得特判一下m == 1即一个人的情况。

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>#include<vector>#include<algorithm>#include<map>#include<set>#include<queue>using namespace std;const int maxn = 20 + 5;#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x )typedef long long ll;#define eps 10e-8const int Mod = 1e9 + 7;typedef pair<int, int> P;ll pows(ll x,ll n){    ll ret = 1;    while(n)    {        if(n & 1)ret = ret * x % Mod;        n >>= 1;x = x * x % Mod;    }    return ret;}ll n,m,k;void solve(){    if(n - 1 - m * k < m - 1)    {        puts("0");return;    }    ll ans = 1;    for(int i = 1; i <= m - 1; i ++)    {        ans = ans * (n - 1 - m * k - i + 1) % Mod * pows(i,Mod - 2) % Mod;    }    ans = ans * n % Mod * pows(m,Mod - 2) % Mod;    printf("%I64d\n",(ans + Mod ) % Mod );}int main(){//    freopen("out.txt","r",stdin);    int Tcase;scanf("%d",&Tcase);    while( Tcase --)    {        scanf("%I64d%I64d%I64d",&n,&m,&k);        if(m == 1)        {            printf("%I64d\n",n);continue;        }        solve();    }    return 0;}

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