【HDOJ5976】Detachment（贪心）

题意：给定n，要求构造若干个各不相同且和为n的正整数使得它们的乘积最大

T<=1e6，1<=n<=1e9

思路：From https://blog.csdn.net/qq_34374664/article/details/53466435

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#include<cmath>
typedef long long ll;
using namespace std;
#define N   110000
#define oo  10000000
#define MOD 1000000007

ll fac[N],inv[N],exf[N],sum[N];

int main()
{ 
    int cas;
    scanf("%d",&cas);
    fac[0]=fac[1]=inv[0]=inv[1]=exf[0]=exf[1]=1;
    sum[1]=0;
    for(int i=2;i<N;i++)
    {
        fac[i]=fac[i-1]*i%MOD;
        inv[i]=inv[MOD%i]*(MOD-MOD/i)%MOD;
        exf[i]=exf[i-1]*inv[i]%MOD;
        sum[i]=sum[i-1]+i;
    }
    while(cas--)
    {
        ll n;
        scanf("%I64d",&n);
        if(n==1)
        {
            printf("1
"); 
            continue;
        }
        int l=2; 
        int r=N-1;
        int last=2;
        while(l<=r)
        {
            int mid=(l+r)>>1;
            if(sum[mid]<=n){last=mid; l=mid+1;}
             else r=mid-1;
        }
        int res=n-sum[last];
        ll ans=0;
        if(res==last) ans=fac[last]*inv[2]%MOD*(res+2)%MOD; //全部+1，剩下的1加在最后一个数上 
         else ans=fac[last+1]*exf[last+1-res]%MOD*fac[last-res]%MOD; //从后往前+1 
        printf("%I64d
",ans);
    }
    return 0;
}