1 solutions

  • 0
    @ 2025-11-5 19:36:11

    C :

    #include<stdio.h>
    #include<math.h>
    long long jc(int n)
    {
    	long long result;
    	int i;
    	for(i=1,result=1;i<=n;i++)
    	result*=i;
    	return result;
    }
    main()
    {
    	long long sum;
    	int n,m,i;
    	scanf("%d",&n);
    	while(n--)
    	{
    		scanf("%d",&m);
    		sum=0;
    		for(int i=1;i<=m-1;i++)
    		sum+=(i+1)*(jc(m-1)/(jc(m-1-i)*jc(i)));
    		printf("%lld\n",sum+1);
    	}
    }
    

    C++ :

    #include <stdio.h>
    int main()
    {
    	int t , num[22] , base = 1 ,i ,n;
    	scanf("%d",&t);
    	num[1] = 1;
    	for(i = 2 ; i<= 20 ; i++)
    	{
    		num[i] = 2 * num[i - 1] + base ;
    		base <<= 1 ;
    	}
    	while(t--)
    	{
    		scanf("%d",&n);
    		printf("%d\n",num[n]);
    	}
    	return 0;
    }
    
    • 1

    Information

    ID
    19441
    Time
    1000ms
    Memory
    128MiB
    Difficulty
    (None)
    Tags
    # Submissions
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