ＴＣSRM 591 div2（１０００）（ｄｐ）

挺好的ｄｐ　因为有一点限制　必须任意去除一个数　总和就会小于另一个总和　换句话来说就是去除最小的满足　那么就都满足

所以是限制最小值的背包　刚开始从小到大定住最小值来背　ＴＬＥ了一组数据　后来发现如果从大到小的话　就不用多加一重ｆｏｒ了　　前面算的已经记录下来　直接用就ＯＫ了　

 #include <iostream>
 #include<cstdio>
 #include<cstring>
 #include<algorithm>
 #include<stdlib.h>
 #include<vector>
 using namespace std;
 #define LL long long
 #define N 3000010
 LL dp[N],sum[N];
 int a[],b[];
 bool cmp(int a,int b)
 {
     return a>b;
 }
 class YetAnotherTwoTeamsProblem
 {
     public:
     long long count(vector <int> skill)
     {
         int i,j,k,n;
         for(i =  ; i < (int)skill.size() ; i++)
         a[i+] = skill[i];
         LL ss=,s=;
         n = (int)skill.size();
         for(i =  ;i <= n ;i++)
         {
             s+=a[i];
         }
         sort(a+,a+n+,cmp);
         for(i =  ; i <= n ;i++)
         sum[i]=sum[i-]+a[i];
         LL v = (s+)/;
         LL vv = s/;
         dp[] = ;
         for(i = ; i < n ;i++)
         {
             for(k = v- ; k >= a[i] ; k--)
             {
                 dp[k]+=dp[k-a[i]];
             }
             if(a[i]>vv)
             ss+=;
             int tt;
             if(vv-a[i+]+<)
             tt = ;
             else
             tt = vv-a[i+]+;
             for(j = tt ; j < v ;j++)
             ss+=dp[j];
         }
         return ss;
     }
 };