Bessie has gone to the mall's jewelry store and spies a charm bracelet. Of course, she'd like to fill it with the best charms possible from the N (1 ≤ N ≤ 3,402) available charms. Each charm i in the supplied list has a weight W_i (1 ≤ W_i ≤ 400), a 'desirability' factor D_i (1 ≤ D_i ≤ 100), and can be used at most once. Bessie can only support a charm bracelet whose weight is no more than M (1 ≤ M ≤ 12,880).

Given that weight limit as a constraint and a list of the charms with their weights and desirability rating, deduce the maximum possible sum of ratings.

* Line 1: Two space-separated integers: N and M
* Lines 2..N+1: Line i+1 describes charm i with two space-separated integers: W_i and D_i

* Line 1: A single integer that is the greatest sum of charm desirabilities that can be achieved given the weight constraints

问题分析：

N 个物品每个物品有价值v[i]，重量w[i]， 给定背包最大承重M，求背包能够装载的最大价值。每个物品只有放入背包和不放入背包两种选择。

这是典型的0-1背包问题。

代码的时间上限是O(nm), 对于每个物品i, 它所要遍历的整数区间都是[ci, m]

#include<iostream>

#include<algorithm>

#include<cstring>

#include<string>

#include<cmath>

using namespace std;

int dp[];

int w[];

int v[];

int main()

{

    int n,m;

    cin>>n>>m;

    memset(dp,,sizeof(dp));

    for(int i=;i<=n;i++)

    {

        cin>>w[i]>>v[i];

    }

    for(int i=;i<=n;i++)

    {

        for(int j=m;j>=w[i];j--)//要倒着推过来，因为dp【j】是滚动数组

        {

            dp[j]=max(dp[j],dp[j-w[i]]+v[i]);

        }

    }

    cout<<dp[m];

    return ;

}