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