HDU1258 Sum it up

Sum it up

题意:给定一个数sum,和n个数,求sum可以由这n个数里面的那几个数的和表示。

Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t=4, n=6, and the list is [4,3,2,2,1,1], then there are four different sums that equal 4: 4,3+1,2+2, and 2+1+1.(A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.

注意:输入,输出要求比较高,另外不能有重复。

​ 递归的时候应该及时退出。

http://acm.hdu.edu.cn/showproblem.php?pid=1258

#include<cstdio>
#include<iostream>
int sum,n;
int flag=0;
int a[20],ans[20];
void  dfs(int sums,int cut,int x)//sums代表当前的和,cut代表ans里的[1,cut),x代表a中的第x个数
{
    if(sums==sum){
        for(int i=1;i<cut;i++){
            flag=1;
            if(i==cut-1)
                printf("%d
",ans[i]);
            else
                printf("%d+",ans[i]);
        }
        return ;
    }//如果结果sums==sum按格式输出ans 并且flag=1;
    int t=-1;
    for(int i=x;i<=n;i++){

        if(t!=a[i]){
            ans[cut]=a[i];
            t=a[i];//避免重复
            dfs(sums+a[i],cut+1,i+1);
        }
    }//
    return ;//递归要有结束条件,不能是return;
}
int main ()
{
    while(scanf("%d %d",&sum,&n),n||sum)
    {
        for(int i=1;i<=n;i++)
            scanf("%d",&a[i]);
        flag=0;
        printf("Sums of %d:
",sum);
        dfs(0,1,1);
        if(flag==0)
            printf("NONE
");
    }
    return 0;
}
想的太多,做的太少。
原文地址:https://www.cnblogs.com/pealicx/p/6115624.html