HDU 1171【01背包水题】
来源:互联网 发布:09总决赛科比每场数据 编辑:程序博客网 时间:2024/06/05 03:44
先上题
Big Event in HDU
Time Limit: 10000/5000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 43557 Accepted Submission(s): 14972
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).
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.
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#include<iostream>using namespace std;#include<algorithm>#include<string.h>int value[5010];int dp[260000];int main(){int n;while (scanf("%d", &n) != EOF, n >0){memset(value, 0, sizeof(value));memset(dp, 0, sizeof(dp));int sum = 0;int j = 0;for (int i = 0; i < n; i++){int a, b;scanf("%d%d", &a, &b);while (b--){value[j++] = a;sum += a;}}for (int i = 0; i < j; i++){for (int j = sum / 2; j >= value[i]; j--){dp[j] = max(dp[j], dp[j - value[i]] + value[i]);}}cout << sum - dp[sum / 2] << " " << dp[sum / 2] << endl;}return 0;}
注意所开的value和dp数组的大小,不要过小,要用memset,={0}不知道为什么会WA,输入n时,要写n>0,不要写,n!=-1,会TLE,被坑了n多次
阅读全文
1 0
- HDU 1171【01背包水题】
- HDU 1171 01背包
- HDU 1171(01背包)
- 01背包水题 HDU 2546饭卡
- HDU 2546饭卡 【01背包水题】
- hdu 1171 Big Event in HDU(01背包&多重背包)
- hdu 1171 01背包变形
- HDU 1171 01背包变形
- HDU 1171 Big Event in HDU (由01背包演变的水题)
- 01背包+完全背包 HDU
- hdu 2844 多重背包模板题 01背包、完全背包、多重背包模板
- HDU—1171 DP 01背包
- 01背包——hdu 1171
- HDU-1171 多重背包
- 【多重背包】HDU 1171
- HDU 1171 背包问题
- hdu 1171 多重背包
- hdu 1171 多重背包
- 在 Linux 上配置 mongodb
- ORACLE RAC心跳&&网络
- Centos7.0 关闭防火墙、更改主机名、SSH免密登录
- Android ListView简单示例
- 排队系统拥塞控制的位置
- HDU 1171【01背包水题】
- JZOJ1383. 奇怪的问题 (2017.8B组)
- 如何正确的更好的停止一个线程?
- Elasticsearch实现原理分析-2
- Mybatis映射器
- 学习路线
- [Poi0202]Travelling Salesman 最近公共祖先
- HDU6118-度度熊的交易计划(最小可行流)
- css3动画效果 正方体3d旋转