HDU1171-Big Event in HDU
2016-09-11 16:51
375 查看
Big Event in HDU
Time Limit: 10000/5000 MS (Java/Others) Memory Limit:65536/32768 K (Java/Others)
Total Submission(s): 36398 Accepted Submission(s): 12639
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
Author
lcy
#include <iostream>
#include <algorithm>
#include <stdio.h>
#include <queue>
#include <string.h>
using namespace std;
int x[5010];
int dp[255555];
int main()
{
int n;
while(~scanf("%d",&n)&&n)
{
memset(dp,0,sizeof dp);
memset(x,0,sizeof x);
int a,b;
int k=0,sum=0;
for(int i=0;i<n;i++)
{
scanf("%d %d",&a,&b);
while(b--)
{
x[k++]=a;sum+=a;
}
}
for(int i=0;i<k;i++)
{
for(int j=sum/2;j>=x[i];j--)
dp[j]=max(dp[j],dp[j-x[i]]+x[i]);
}
printf("%d %d\n",sum-dp[sum/2],dp[sum/2]);
}
return 0;
}
相关文章推荐
- HDU 5873 Football Games 【模拟】 (2016 ACM/ICPC Asia Regional Dalian Online)
- 【CodeVS 1218】【NOIP 2012】疫情控制
- Timus Online Judge:2010. Sasha the Young Grandmaster
- ubuntu14安装nodejs
- 操作系统μC/OS-Ⅱ读书笔记(2)
- 设计模式之观察者模式
- django(8):简单配置admin页
- jdbc连接池详解
- apple food
- NYOJ题目893十字架
- HDU2191-悼念512汶川大地震遇难同胞——珍惜现在,感恩生活
- oracle 配置服务端
- Java序列化技术
- MFC----GetDocument()使用
- 向量的基础运算
- STC12C5A32S2单片机之1602渐入显示
- synchronized、volatile
- Poedu_C语言_lesson12_20160911_小数
- 并查集(disjoint set)的实现及应用
- SVN配置出现svnserve.conf:12错误解决