CodeChef - NWAYS 组合数 朱世杰恒等式

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题意:求\(\sum_{i=1}^n\sum_{j=1}^n{|i-j|+k \choose k}\)

知识点: 朱世杰恒等式,\(\sum_{i=r}^n{i \choose r}={n+1 \choose r+1},r<n\)

题解:首先去除式子中的绝对值,考虑对称性还有i=j时的重复,原式可转化为\(2\sum_{i=1}^n\sum_{j=i}^n{j-i+k \choose k}-n\)

对式子内部循环调用一遍朱世杰恒等式\(\sum_{j=i}^n{j-i+k \choose k}=\sum_{j=k}^{k+n-i}{j \choose k}={{k+n-i+1} \choose {k+1}}\) (对中间式子有疑惑的可自行展开)

再对外部循环调用一遍\(\sum_{i=1}^n{{k+n-i+1} \choose {k+1}}=\sum_{i=k+1}^{k+n}{i \choose {k+1}}={{k+n+1} \choose {k+2}}\)

福利:\(\sum_{i=m}^n{i \choose r}={n+1 \choose r+1}-{m \choose r+1}\)

#include<bits/stdc++.h>
#define rep(i,j,k) for(int i=j;i<=k;i++)
using namespace std;
typedef long long ll;
const ll mod = 1000000007;
const int maxn = 2e6+111;////
ll jie[maxn],inv[maxn];
ll fpw(ll a,ll n){
    ll ans=1;
    while(n){
        if(n&1) ans=(ans*a)%mod;
        n>>=1; a=(a*a)%mod;
    }
    return ans;
}
ll C(ll n,ll k){
    ll up=jie[n];
    ll down=inv[k]*inv[n-k]%mod;
    return (up*down)%mod;
}
int main(){
    ll T; scanf("%lld",&T);
    jie[0]=inv[0]=1;
    rep(i,1,maxn-2) jie[i]=(jie[i-1]*i)%mod;
    rep(i,1,maxn-2) inv[i]=fpw(jie[i],mod-2);
    while(T--){
        ll n,k; scanf("%lld%lld",&n,&k);
        ll tmp=C(n+k+1,k+2);
        ll ans=((tmp*2)%mod-n+mod)%mod;
        printf("%lld\n",ans);
    }
    return 0;
}

原文地址：https://www.cnblogs.com/caturra/p/8454713.html