LightOJ 1282 - Leading and Trailing（快速幂取模）

最新推荐文章于 2021-06-12 20:01:56 发布

老焦哥

最新推荐文章于 2021-06-12 20:01:56 发布

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分类专栏：数论

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Description
You are given two integers: n and k, your task is to find the most significant three digits, and least significant three digits of nk.

Input
Input starts with an integer T (≤ 1000), denoting the number of test cases.

Each case starts with a line containing two integers: n (2 ≤ n < 231) and k (1 ≤ k ≤ 107).

Output
For each case, print the case number and the three leading digits (most significant) and three trailing digits (least significant). You can assume that the input is given such that nk contains at least six digits.

Sample Input
5
123456 1
123456 2
2 31
2 32
29 8751919
Sample Output
Case 1: 123 456
Case 2: 152 936
Case 3: 214 648
Case 4: 429 296

Case 5: 665 669

#include <stdio.h>
#include <cmath>
const int mod = 1000;

int powermod(int a, int b)
{
	int ans = 1;
	a %= mod;
	while (b)
	{
		if (b & 1)
			ans = (ans*a) % mod;
		a = (a*a) % mod;
		b /= 2;
	}
	return ans;
}

int main()
{
	int T, cc = 1;
	scanf("%d", &T);
	while (T--)
	{
		int n, k;
		int ans1, ans2;
		scanf("%d%d", &n, &k);
		ans1 = (int)pow(10.0, 2.0 + fmod(k*log10(n), 1));
		ans2 = powermod(n, k);
		printf("Case %d: %d %03d\n", cc++, ans1, ans2);
	}
	return 0;
}