快速幂被卡两个点

#include <bits/stdc++.h>
/*
#include<ext/pb_ds/assoc_container.hpp>
#include<ext/pb_ds/hash_policy.hpp>
*/
using namespace std;

const double eps = 1e-10;
const double pi = 3.1415926535897932384626433832795;
const double eln = 2.718281828459045235360287471352;

#define f(i, a, b) for (int i = a; i <= b; i++)
#define scan(x) scanf("%d", &x)
#define mp make_pair
#define pb push_back
#define lowbit(x) (x&(-x))

#define fi first
#define se second
#define SZ(x) int((x).size())
#define all(x) x.begin(), x.end()
#define rall(x) x.rbegin(), x.rend()
#define summ(a) (accumulate(all(a), 0ll))

typedef unsigned long long ull;
typedef pair<int,int> pii;
typedef vector<int> vi;

using ll=long long;

const int mod=1e9+7;
ll n,m,c,ans;

int qpow(int a,int b){
    int ret=1;
    for(;b;b>>=1,a=1ll*a*a%mod)if(b&1)ret=1ll*ret*a%mod;
    return ret;
}
int inv(int x){return qpow(x,mod-2);}
int fact[(int)1e5+9],inv_fact[(int)1e5+9],qow[(int)2e5+9]={1};
int C(int n,int m){
    if(m>n||n<0||m<0)return 0;
    return 1ll*fact[n]*inv_fact[m]%mod*inv_fact[n-m]%mod;
}

int main()
{
    ios::sync_with_stdio(false);
    cin.tie(0);
    fact[0]=1;
    f(i,1,400)fact[i]=1ll*fact[i-1]*i%mod;
    inv_fact[400]=inv(fact[400]);
    for(int i=400-1;i>=0;i--)inv_fact[i]=1ll*inv_fact[i+1]*(i+1)%mod;

    cin>>n>>m>>c;

    for(int k=0;k<=c;++k){
        for(int i=1;i<=n*m;++i)qow[i]=qow[i-1]*(c-k+1)%mod;
        for(int i=0;i<=n;++i)for(int j=0;j<=m;++j)ans=(ans+((i+j+k)&1?-1ll:1ll)*C(n,i)*C(m,j)%mod*C(c,k)%mod*qow[(n-i)*(m-j)]%mod+mod)%mod;
    }
    cout<<ans;
    return 0;
}
此文章已被阅读次数:正在加载...更新于