hdu1060 Leftmost Digit

hdu1060 Leftmost Digit

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 14951    Accepted Submission(s): 5774

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

Hint

Author

Ignatius.L

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说实话 这道题我不会做、、当我看到n的范围那么大 就想着会不会有规律。可n又太大。不知道怎么办

就百度了

m=n^n,两边同时取对数log10,变为log10(m)=n*log10(n);继续化简为m=10^(n*log10(n)),因为10的任意整数次方

首位都是1，于是首位就和n*log10(n)的小数部分有关了。

还有这样的方法对于2来说是不对的。。不

#include #include int main(){ int ncase; double n,ans; scanf("%d",&ncase); while(ncase--) { scanf("%lf",&n); ans=log10(n); ans=(ans-(int)ans)*n; ans=ans-(int)ans; printf("%d\n",(int)pow(10,ans)); } return 0;}

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