【杭電-oj】 -1060-Leftmost Digit(輸出n的n次方最左邊數)
阿新 • • 發佈:2019-01-30
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.
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
n^n=a*10^m;
兩邊對10去對數,得n*lg(n)=m+lg(a);
此處接下來用一個數學的pow,例如pow(n,m)表示n^m;
因為1<a<10,所以0<lg(a)<1.
接下來還有一個數學知識,即x^log(x)(n)=n;
#include<stdio.h> #include<math.h> int main() { int t,ans; double n; scanf("%d",&t); while(t--) { scanf("%lf",&n); double x,a; x=n*log10(n); a=pow(10,x-(__int64)x); //x-(__int64)x,表示n*lgn的小數部分,即lg(a) ans=int(a); printf("%d\n",ans); } return 0; }