POJ 1564 经典dfs

1、POJ 1564 Sum It Up

2、总结：

题意：在n个数里输出所有相加为t的情况。

#include<iostream>
#include<cstring>
#include<cmath>
#include<queue>
#include<algorithm>
#include<cstdio>
#define F(i,a,b) for (int i=a;i<=b;i++)
#define mes(a,b) memset(a,b,sizeof(a))
#define INF 0x3f3f3f3f
#define LL long long
using namespace std;
const int N=10010,MAX=1000100;

bool flag;
int t,n;
int num[20],ans[20];

void dfs(int cnt,int k,int sum)
{
    if(sum==t){
        flag=false;
        F(i,1,cnt-1){
            if(i==1)printf("%d",ans[i]);
            else printf("+%d",ans[i]);
        }
        printf("\n");
        return ;
    }

    //关键
    F(i,k,n){
        if(i==k||num[i]!=num[i-1]&&sum+num[i]<=t){
            ans[cnt]=num[i];
            dfs(cnt+1,i+1,sum+num[i]);
        }
    }
}

int main()
{
    while(~scanf("%d%d",&t,&n),t&&n)
    {
        flag=true;
        F(i,1,n)scanf("%d",&num[i]);
        printf("Sums of %d:\n",t);
        dfs(1,1,0);
        if(flag)printf("NONE\n");

    }

    return 0;
}