逆元，排列组合

题目链接：传送门

题意：
求解构造的序列（本质不同）的贡献和
代码如下：

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll maxn=1e6+5;
const ll MOD=998244353;
ll n,m,ans,jie[maxn];
ll quick(ll a,ll b){
    ll ret=1;
    while(b){
        if(b&1)ret=ret*a%MOD;
        a=a*a%MOD,b>>=1;
    }
    return ret;
}
ll f(ll _n,ll _m){
   ll mu;
   mu=jie[_m]*jie[_n-_m]%MOD;
   return jie[_n]*quick(mu,MOD-2)%MOD;
}
int main(){
    ll i,j,tmp;
    jie[0]=jie[1]=1;
    for (i=2;i<maxn;i++)jie[i]=jie[i-1]*i%MOD;
    while(scanf("%lld%lld",&n,&m)!=EOF){
        ans=0;j=m-1;
        for (i=1;i<n;i++){
            tmp=i*(n-i)%MOD;tmp=tmp*f(j,i)%MOD;ans=(ans+tmp)%MOD;
            j++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}

标签：tmp,jie,ll,ret,逆元,ans,排列组合,MOD
来源： https://blog.csdn.net/qq_39507939/article/details/114145695