hdu-1171 Big Event in HDU（01背包）

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44492 Accepted Submission(s): 15311

Problem Description
Nowadays, we all know that Computer College is the biggest department in HDU. But, maybe you don’t know that Computer College had ever been split into Computer College and Software College in 2002.
The splitting is absolutely a big event in HDU! At the same time, it is a trouble thing too. All facilities must go halves. First, all facilities are assessed, and two facilities are thought to be same if they have the same value. It is assumed that there is N ( 0< N < 1000 ) kinds of facilities (different value, different kinds).

Input
Input contains multiple test cases. Each test case starts with a number N ( 0 < N <= 50 – the total number of different facilities). The next N lines contain an integer V ( 0 < V<=50 –value of facility) and an integer M (0 < M<=100 –corresponding number of the facilities) each. You can assume that all V are different.
A test case starting with a negative integer terminates input and this test case is not to be processed.

Output
For each case, print one line containing two integers A and B which denote the value of Computer College and Software College will get respectively. A and B should be as equal as possible. At the same time, you should guarantee that A is not less than B.

Sample Input
2
10 1
20 1
3
10 1
20 2
30 1
-1

Sample Output
20 10
40 40

emmmmmmmm， 没看清题， 看数据-1 结束 解这样以为了 超时了
其实是>0,
这道题可用01背包解决， 把多个数拆开存入数组 当一个一个就好。
一般都是有包的容量和和物品信息， 这道题的包的容量就相当于每个物品的价值和 sum

如果想要满足 a和b 尽量接近， 我们需要达到的最大 价值和为 sum/2,

其他就和01背包差不多一样了

code：

#include <stdio.h>#include <iostream>#include <bits/stdc++.h>using namespace std;int value[10003];int dp[100003];int main(){    int n;    while (scanf ("%d",&n),n>0)    {        memset(dp, 0, sizeof(dp));        memset(value, 0, sizeof(value));        int k=0, sum=0;        int x, m;        for (int i=0; i<n; i++)        {            scanf ("%d%d",&x, &m);            for (int j=0; j<m; j++)            {                value[k++] = x;                sum += x;            }        }        for (int i=0; i<k; i++)        {            for (int j=sum/2; j>=value[i]; j--)            {                dp[j] = max(dp[j], dp[j-value[i]]+value[i]);            }        }        printf ("%d %d\n",sum-dp[sum/2], dp[sum/2]);    }    return 0;}