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  • HDU 1060

    Leftmost Digit

    Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other)
    Total Submission(s) : 7   Accepted Submission(s) : 2

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    Problem Description

    Given a positive integer N, you should output the leftmost digit of N^N.

    Input

    The input contains several test cases. The first line of the input is a single integer T which is the number of test cases. T test cases follow.
    Each test case contains a single positive integer N(1<=N<=1,000,000,000).

    Output

    For each test case, you should output the leftmost digit of N^N.

    Sample Input

    2
    3
    4
    

    Sample Output

    2
    2
    

    Hint

    In the first case, 3 * 3 * 3 = 27, so the leftmost digit is 2.
    In the second case, 4 * 4 * 4 * 4 = 256, so the leftmost digit is 2.


    题意:

      给你一个数字n。输出n的n次的第一位。

    题解:

      n^n=n*log10(n)。而10的整数次方数的首位为1,因此要求此题的答案仅仅需求出n*log10(n)的小数部分m。由于0<m<10,所以10^m的整数部分((int)pow(10,m))即为答案。

    參考代码:

    #include<stdio.h>
    #include<math.h>
    int main()
    {
    	int t;
    	scanf("%d",&t);
    	while(t--)
    	{
    		double a,c,n;
    		long long d,b;
    		scanf("%lf",&n);
    		a=n*log10(n);
    		b=(long long)a;//因为题目数据量的限制。这里的b的强制转换为long long型。int型会wa。
    		c=a-b;
    		d=(int)pow(10.0,c);
    		printf("%lld
    ",d);
    	}
    	return 0;
    }

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  • 原文地址:https://www.cnblogs.com/zhchoutai/p/6699496.html
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