Leftmost Digit |
| Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others) |
| Total Submission(s): 364 Accepted Submission(s): 198 |
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Problem Description
Given a positive integer N, you should output the leftmost digit of N^N.
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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). |
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Output
For each test case, you should output the leftmost digit of N^N.
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Sample Input
2 3 4 |
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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. |
题目大意是输入N，求N^N的最高位数字。1<=N<=1,000,000,000
估计大家看到N的范围就没想法了。确实N的数字太大，如果想算出结果，即使不溢出也会超时。
这题我纠结了很久。在同学的提示下ac了。
题目是这样转化的。
首先用科学计数法来表示 N^N = a*10^x; 比如N = 3; 3^3 = 2.7 * 10^1;
我们要求的最右边的数字就是(int)a，即a的整数部分;
OK, 然后两边同时取以10为底的对数 lg(N^N) = lg(a*10^x) ;
化简 N*lg(N) = lg(a) + x;
继续化 N*lg(N) - x = lg(a)
a = 10^(N*lg(N) - x);
现在就只有x是未知的了，如果能用n来表示x的话，这题就解出来了。
又因为，x是N^N的位数。比如 N^N = 1200 ==> x = 3; 实际上就是 x 就是 lg(N^N) 向下取整数，表示为[lg(N^N)]
ok a = 10^(N*lg(N) - [lg(N^N)]); 然后(int)a 就是答案了。
代码:
#include<iostream>
#include<math.h>
int main()
{
int n,m;
std::cin>>n;
while(n--)
{
std::cin>>m;
long double t = m*log10(m*1.0);
t -= (__int64)t;
__int64 ans = pow((long double)10, t);
std::cout<<ans<<std::endl;
}
return 0;
}