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  • Trailing Zeroes (III)

    Description

    You task is to find minimal natural number N, so that N! contains exactly Q zeroes on the trail in decimal notation. As you know N! = 1*2*...*N. For example, 5! = 120, 120 contains one zero on the trail.

    Input

    Input starts with an integer T (≤ 10000), denoting the number of test cases.

    Each case contains an integer Q (1 ≤ Q ≤ 108) in a line.

    Output

    For each case, print the case number and N. If no solution is found then print 'impossible'.

    Sample Input

    3

    1

    2

    5

    Sample Output

    Case 1: 5

    Case 2: 10

    Case 3: impossible

    <pre name="code" class="cpp">#include<cstdio>
    long long  qiuling(long long p)//球p的阶乘尾部有几个零 
    {
    	long long sum=0;
        while(p)
        {
        	sum+=p/5;
        	p/=5;
    	}
    	return sum;
    }
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	long long  cut=0,q;
    	while(t--)
    	{
    		cut++;
    		scanf("%lld",&q);
            long long zuo=1,you=900000000,ans,mid;
            while(zuo<=you)//用二分查找
            {
            	mid=(zuo+you)/2;
            	if(qiuling(mid)>=q)
            	{
            		ans=mid;//必须定义一个数来记录mid   如果最后直接输出mid 虽然测试正确但不能AC 
            		you=mid-1;
    			}
    			else
    			{
    				zuo=mid+1;
    			}
    		}
    		if(qiuling(ans)!=q)
    		{
    			printf("Case %lld: impossible
    ",cut);
    		}
    		else
    		{
    			printf("Case %lld: %lld
    ",cut,ans);
    		}
    	}
      return 0;	
     } 


    
    
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  • 原文地址:https://www.cnblogs.com/kingjordan/p/12027139.html
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