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[55],b[55];
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 = 0 ; i < (int)skill.size() ; i++)
        a[i+1] = skill[i];
        LL ss=0,s=0;
        n = (int)skill.size();
        for(i = 1 ;i <= n ;i++)
        {
            s+=a[i];
        }
        sort(a+1,a+n+1,cmp);
        for(i = 1 ; i <= n ;i++)
        sum[i]=sum[i-1]+a[i];
        LL v = (s+1)/2;
        LL vv = s/2;
        dp[0] = 1;
        for(i = 1; i < n ;i++)
        {
            for(k = v-1 ; k >= a[i] ; k--)
            {
                dp[k]+=dp[k-a[i]];
            }
            if(a[i]>vv)
            ss+=1;
            int tt;
            if(vv-a[i+1]+1<0)
            tt = 0;
            else
            tt = vv-a[i+1]+1;
            for(j = tt ; j < v ;j++)
            ss+=dp[j];
        }
        return ss;
    }
};