dividing coins 简单dp

最新推荐文章于 2020-02-25 19:29:32 发布

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本博客介绍了一个算法，用于解析输入中的硬币数量和价值，并计算将这些硬币分成两组时，两组之间的最小价值差异。通过迭代过程找到最接近总和一半的组合，进而计算差异。

Input

A line with the number of problems n, followed by n times:

a line with a non negative integer m ( $m \le 100$ ) 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.

#include<stdio.h>
#include<string.h>
int n,a[110];
int dp[300010];
int main()
{
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%d",&n);
        int sum=0;
        for(int i=0;i<n;i++){
            scanf("%d",&a[i]);
            sum+=a[i];
        }
        memset(dp,0,sizeof(dp));
        dp[0]=1;
        int mid=sum/2;
        for(int i=0;i<n;i++)
        {
            for(int j=mid;j>=a[i];j--)  // for(int j=a[i];j<=mid;j++) 这是有本质区别！倒着循环a[i]只用了一次，正着循环表示可以使用mid-a[i] 次
            {
                if(dp[j-a[i]]) dp[j]=1;
            }
            if(dp[mid]) break;
        }
        for(;mid>=0;mid--)if(dp[mid])
        {
            printf("%d\n",sum-2*mid);
            break;
        }
    }
}