HDU--航但--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): 20769    Accepted Submission(s): 7268

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

210 120 1310 1 20 230 1-1

 

Sample Output

20 1040 40

 

题意：告诉你有n组数据，每组是一个值a和这个值可用的个数b，把他们分成差值最小的两个数字

题解：要么是一个大一个小，要么就是两个相同，所有那个较小的数字最大只是总和的一半，用总和的一半来01背包就可以解了

#include <iostream>using namespace std;int main (void){    int n,i,j,k,l,a[55],b[55],dp[255555],x;    while(cin>>n&&n>0)    {        for(i=0;i<255555;i++)dp[i]=0;        dp[0]=1;        for(i=x=0;i<n;i++)        {            cin>>a[i]>>b[i];            x+=a[i]*b[i];//总和        }        for(i=0;i<n;i++)        for(j=1;j<=b[i];j++)        for(k=x/2;k>=a[i]*j;k--)        if(dp[k-a[i]*j])dp[k]=1;        for(i=x/2;i>0;i--)if(dp[i])break;//把最大的那个较小数拿出来        cout<<x-i<<" "<<i<<endl;//(A+B)-A=B    }    return 0;}