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\n",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:\n",sum);
dfs(0,1,1);
if(flag==0)
printf("NONE\n");
}
return 0;
}

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