Leftmost Digit（杭电1060）（求N^N的最高位）

Leftmost Digit

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 13574    Accepted Submission(s): 5216

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

234

 

Sample Output

22
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.
 

 

Author

Ignatius.L

/*本题思路： 1,令M=N^N； 2，分别对等式两边取对数得 log10(M)=N*log10(N),得M=10^(N*log10(N))；                  3，令N*log10(N)=a+b,a为整数，b为小数；4，C函数：log10()，计算对数，pow(a,b)计算a^b5，由于10的任何整数次幂首位一定为1,所以,M的首位只和N*log10(N)的小数部分有关，   即只用求10^b救可以了；   6，最后对10^b取整，输出取整的这个数就行了。（因为0<=b<1,所以1<=10^b<10对其取整，那么的到的就是一个个位，也就是所求的数）。*///hdoj系统就是坑，各种CE，各种WA，多亏康晓辉指点。 #include<stdio.h>#include<math.h>int main(){int test;double n,k,a;scanf("%d",&test);while(test--){scanf("%lf",&n);a=n*log10(n);k=a-(__int64)a; //这里强制转换需要用int64才能A，否则会w或者CE！！！ printf("%d\n",int(pow(10,k)));}return 0;}