HDU 1171 Big Event in HDU（母函数||DP）

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 22991    Accepted Submission(s): 8079

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

 

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 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

 WA了 好多次  就以为结束条件写的 while(cin >> n && n!=-1)   它要是提示时间超限也行呀。。
无语。。不吐了，进入正题
题意： 杭电本来只有一个计算机院，但是要分出来一个软件院 
 就相当与兄弟两个要分家了，家产也应该分了。
输入 n
n行设备， 第一个数是价值，第一个数是数量。
输出：一行，计算机院和软件院最终所得价值。（能平分是最好，前提设备不能被破坏）不能正好平分就尽可能的保证公平，计算机院的要多一些
思路：
用母函数模板进行一些变形
AC代码

#include 
#include 
#include 
#include 
#include 
#define MAXN 1001000
using namespace std;

int value[200],number[200],ans[MAXN],temp[MAXN];
int main()
{
    int n,i,j,k;
    while(cin >> n&&n>=0)
    {
        int max = 0;
        for(i=0; i> value[i] >> number[i];
            max += value[i] * number[i]; //设备的总价值
        }
        int mid = max/2 ;   //情况最好的时候是平分 即mid
        memset(ans,0,sizeof(int)*mid+10);
        memset(temp,0,sizeof(int)*mid+10);
        for(i=0; i<=number[0]; i++)     //用母函数数法 把第一排赋初值
            ans[i*value[0]] = 1;
        for(i=1; i=0; i--)
            if(ans[i]!=0)
                break;
        cout << max - i<<" "<< i << endl;
    }
    return 0;
}