1 solutions

  • 0
    @ 2025-3-3 16:31:20

    C :

    #include <stdio.h>
    long int a[30];
    long int b[30];
    long int T;
    int n;
    long int sum=0;
    void f(int long t,int pos,int i )
    {int j;
    
        if(t==T&&pos!=0&&i<1)//终止条件 注意全部遍历完再进行判断(防止 -1 1 -2 2 3 -3的特例) 并且 不能不选 
        {sum++;
        for(j=pos-1;j>=0;j--)//倒着遍历倒着输出 
        printf("%ld ",b[j]);
        printf("\n");
        }
        else
        {  if(i<1)return ;//遍历到数组第一项还未满足t==T;则返回 
    
            f(t,pos,i-1);
             t+=a[i];b[pos]=a[i];
             f(t,pos+1,i-1);
             t-=a[i];
    
    
        }
    
    }
    int main()
    {
    int i;
    scanf("%d",&n);
    for(i=1;i<=n;i++)
    scanf("%ld",&a[i]);
    scanf("%ld",&T);
    f(0,0,n);
    printf("%d\n",sum);
    return 0;
    }
    
    • 1

    Information

    ID
    10346
    Time
    1000ms
    Memory
    512MiB
    Difficulty
    (None)
    Tags
    # Submissions
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