1、   

2、总结：

题意：在n个数里输出所有相加为t的情况。

#include
      
       #include
       
        #include
        
         #include
         
          #include
          
           #include
           
            #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 longusing 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;}