打怪兽 51Nod - 1670

https://www.51nod.com/Challenge/Problem.html#!#problemId=1670

参考http://www.cnblogs.com/whistle13326/p/7739636.html

dp[i]表示第i轮存活下来的希望 可以找到上述博客中 dp[i+1]=dp[i] *(x-(i+1)+1)/(n-(i+1)+1) 这个递推关系 对每一轮可求出至少活到当前轮的概率 那dp[i]-dp[i+1]就代表战胜第i个怪兽 败给第(i+1)个怪兽的概率 乘上系数即可

 

#include <bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll mod=1e9+7;
const int maxn=1e5+10;

ll sum[maxn],pre[maxn],dp[maxn];
int n;

void init()
{
    ll i;
    pre[0]=1;
    for(i=1;i<=1e5;i++) pre[i]=(i*pre[i-1])%mod;
}

ll quickpow(ll a,ll b)
{
    ll res;
    res=1;
    while(b>0){
        if(b%2) res=(res*a)%mod;
        a=(a*a)%mod,b/=2;
    }
    return res;
}

int main()
{
    ll ans,i;
    int tmp;
    init();
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&tmp);
        sum[tmp]++;
    }
    for(i=1;i<=n-1;i++) sum[i]+=sum[i-1];
    dp[0]=pre[n];
    for(i=1;i<=n;i++){
        dp[i]=((dp[i-1]*(sum[i-1]-(i-1))%mod)*quickpow(n-(i-1),mod-2))%mod;
    }
    ans=0;
    for(i=2;i<=n;i++){
        ans=(ans+((dp[i-1]-dp[i]+mod)%mod)*(i-1))%mod;
    }
    ans=(ans+dp[n]*(ll)(n)%mod)%mod;
    printf("%lld\n",ans);
    return 0;
}