HDU

Big Event in HDU

Time Limit: 10000/5000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 44223    Accepted Submission(s): 15217

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：给出一堆物品的价值和数量 把这堆物品分成两小堆 使得两者的价值最接近且第一堆的价值大于等于第二堆的价值

思路：如果一个物品数量大于1 就分成多个相同的物品 用总价值的一半来dp

#include <iostream>#include <cstdio>#include <cstring>#include <algorithm>#define ll long long#define max_ 1400using namespace std;int dp[5010*50],num[5010];int main(){int n;while(scanf("%d",&n)!=EOF&&n>=0){int sum=0,i,j,k,cnt=1;memset(dp,0,sizeof(dp));memset(num,0,sizeof(num));for(i=1;i<=n;i++){int a,b;scanf("%d%d",&a,&b);sum+=a*b;while(b--){num[cnt++]=a;}}for(i=1;i<cnt;i++){for(j=sum/2;j>=num[i];j--)dp[j]=max(dp[j],dp[j-num[i]]+num[i]);}printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);}}