UVA 562 Dividing coins 01背包

把所有数的和的一半看作是背包的容量,然后做一个01背包就可以搞了————题解为数不多看了题解1A的题~~怎么 感觉之前看到了这个呢

#include <iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
using namespace std;
int price[50500],f[50500],n,t,sum;
int dp()
{
     memset(f,0,sizeof(f));
     for(int i=1;i<=n;i++)
     {
          for(int j=sum;j>=price[i];j--)
          {
               if(f[j]<f[j-price[i]]+price[i])
               f[j]=f[j-price[i]]+price[i];
          }
     }
     return f[sum];
}
int main()
{
   // freopen("cin.txt","r",stdin);
    while(~scanf("%d",&t))
    {
         while(t--)
         {
              scanf("%d",&n);
               sum=0;
               for(int i=1;i<=n;i++)
               {
                    scanf("%d",&price[i]);
                    sum+=price[i];
               }
               int cnt=sum;
               sum=(sum+1)/2;
               cnt=dp()*2-cnt;
               if(cnt<0) cnt=-cnt;
               printf("%d\n",cnt);
         }
    }
    return 0;
}

—>_—>

Description

Download as PDF

It's commonly known that the Dutch have invented copper-wire. Two Dutch men were fighting over a nickel, which was made of copper. They were both so eager to get it and the fighting was so fierce, they stretched the coin to great length and thus created copper-wire.


Not commonly known is that the fighting started, after the two Dutch tried to divide a bag with coins between the two of them. The contents of the bag appeared not to be equally divisible. The Dutch of the past couldn't stand the fact that a division should favour one of them and they always wanted a fair share to the very last cent. Nowadays fighting over a single cent will not be seen anymore, but being capable of making an equal division as fair as possible is something that will remain important forever...


That's what this whole problem is about. Not everyone is capable of seeing instantly what's the most fair division of a bag of coins between two persons. Your help is asked to solve this problem.


Given a bag with a maximum of 100 coins, determine the most fair division between two persons. This means that the difference between the amount each person obtains should be minimised. The value of a coin varies from 1 cent to 500 cents. It's not allowed to split a single coin.

Input 

A line with the number of problems  n , followed by  n  times:
  • a line with a non negative integer m () indicating the number of coins in the bag
  • a line with m numbers separated by one space, each number indicates the value of a coin.

Output 

The output consists of  n  lines. Each line contains the minimal positive difference between the amount the two persons obtain when they divide the coins from the corresponding bag.

Sample Input 

2
3
2 3 5
4
1 2 4 6

Sample Output 

0
1

全部评论

相关推荐

本神尊:看来是没招到小红薯上的人
点赞 评论 收藏
分享
06-17 00:26
门头沟学院 Java
程序员小白条:建议换下项目,智能 AI 旅游推荐平台:https://github.com/luoye6/vue3_tourism_frontend 智能 AI 校园二手交易平台:https://github.com/luoye6/vue3_trade_frontend GPT 智能图书馆:https://github.com/luoye6/Vue_BookManageSystem 选项目要选自己能掌握的,然后最好能自己拓展的,分布式这种尽量别去写,不然你只能背八股文了,另外实习的话要多投,尤其是学历不利的情况下,多找几段实习,最好公司title大一点的
无实习如何秋招上岸
点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客网在线编程
牛客网题解
牛客企业服务