Farmer John has a balance for weighing the cows. He also has a set of N (1 <= N <= 1000) weights with known masses (all of which fit in 31 bits) for use on one side of the balance. He places a cow on one side of the balance and then adds weights to the other side until they balance. (FJ cannot put weights on the same side of the balance as the cow, because cows tend to kick weights in his face whenever they can.) The balance has a maximum mass rating and will break if FJ uses more than a certain total mass C (1 <= C < 2^30) on one side. The weights have the curious property that when lined up from smallest to biggest, each weight (from the third one on) has at least as much mass as the previous two combined. FJ wants to determine the maximum mass that he can use his weights to measure exactly. Since the total mass must be no larger than C, he might not be able to put all the weights onto the scale. Write a program that, given a list of weights and the maximum mass the balance can take, will determine the maximum legal mass that he can weigh exactly.

* Line 1: Two space-separated positive integers, N and C.

* Lines 2..N+1: Each line contains a single positive integer that is the mass of one weight. The masses are guaranteed to be in non-decreasing order.

第2到N+1行：每一行仅包含一个正整数，即某个砝码的质量．保证这些砝码的质量是一个不下降序列

* Line 1: A single integer that is the largest mass that can be accurately and safely measured.

一个正整数，表示用所给的砝码能称出的不压坏天平的最大质量．

INPUT DETAILS:

FJ has 3 weights, with masses of 1, 10, and 20 units. He can put at most 15
units on one side of his balance.

约翰有3个砝码，质量分别为1，10，20个单位．他的天平最多只能承受质量为15个单位的物体．用质量为1和10的两个砝码可以称出质量为11的牛．这3个砝码所能组成的其他的质量不是比11小就是会压坏天平

Silver

这道题看到了以为是背包dp；然后突然发现数据那么大Orz就跪了，遂找题解。。。原来还可以这样子搞啊。这怎么想到的呢。反正记住就好了。听说是有技巧的背包。。

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#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
#define ll long long
int n,m;
ll ans=-1;
const int nmax=1010;
int a[nmax];
ll sum[nmax];
void dfs(int cur,ll count){
 if(count>m) return;
 if(sum[cur-1]+count<=m){
  ans=max(ans,sum[cur-1]+count);
  return ;
 }
 ans=max(ans,count);
 for(int i=1;i<cur;i++){
  count+=a[i];
  dfs(i,count);
  count-=a[i];
 }
}
int main(){
 scanf("%d%d",&n,&m);
 for(int i=1;i<=n;i++){
  scanf("%d",&a[i]);
  sum[i]=sum[i-1]+a[i];
 }
 dfs(n+1,0);
 printf("%d\n",ans);
 return 0;
}

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