zoukankan      html  css  js  c++  java
  • HDU-1171

    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).

    InputInput 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.
    OutputFor 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
    
    
     1 #include<iostream>
     2 #include<stdio.h>
     3 #include<string.h>
     4 using namespace std;
     5 int dp[100005];
     6 struct Node{
     7     int val;
     8     int num;
     9 }a[1005];
    10 int main()
    11 {
    12     int n;
    13     while(cin>>n)
    14     {
    15         if(n<0)break;
    16         int sum=0;
    17         memset(dp,0,sizeof(dp));
    18         for(int i=1;i<=n;i++)
    19         {
    20             cin>>a[i].val>>a[i].num;
    21             sum+=a[i].val*a[i].num;
    22         }
    23         int mid=sum/2;
    24         for(int i=1;i<=n;i++)
    25         {
    26             for(int k=1;k<=a[i].num;k++)
    27             {
    28                 for(int j=mid;j>=k*a[i].val;j--)
    29                 {
    30                     dp[j]=max(dp[j],dp[j-k*a[i].val]+k*a[i].val);
    31                 }
    32             }
    33         }
    34         cout<<max(dp[mid],sum-dp[mid])<<' '<<min(dp[mid],sum-dp[mid])<<endl;
    35     }
    36     return 0;
    37 }
    
    
    
    
    
    
    
  • 相关阅读:
    第一次离职了
    华为手机 关于Android真机调试的时候logcat中无法输出调试信息的解决办法
    ListView的简单实用
    博客生涯开始
    hdu1151:Air Raid(最小路径覆盖)
    hdu1151:Air Raid(最小路径覆盖)
    hdu1162:Eddy's picture(最小生成树)
    hdu1162:Eddy's picture(最小生成树)
    hdu1301:Jungle Roads(最小生成树)
    hdu1301:Jungle Roads(最小生成树)
  • 原文地址:https://www.cnblogs.com/ISGuXing/p/7215302.html
Copyright © 2011-2022 走看看