LightOJ 1282 Leading and Trailing（快速幂+对数转换）

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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
题意：给你一个数n，让你求这个数的k次方的前三位和最后三位

后三位可以用快速幂求出来。
前三位就要用到log了。先设10^p =n^k，同时取 log10，那么p=k* log10(n)。再设x=(int)p（整数部分），y=p-x（小数部分），那么10^p = 10^x * 10^ y；由于10^x 是10的倍数，那么10^ y=n^k /10^x，就接近 n^k 的值，只不过小数点在最高位后面，所以还要再乘100.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

int quick_pow(int n, int k){
	int ans=1;
	int base=n%1000;
	while (k){
		if (k&1)
			ans=ans*base%1000;
		base=base*base%1000;
		k>>=1;
	}
	return ans;
}
int main(){
	int n, k, t, cas=1;
	scanf("%d", &t);
	while (t--){
		scanf("%d%d", &n, &k);
		double p=(double)k*log10(n*1.0);
		p-=(int)p;
		double tt=pow(10.0, p);
		printf("Case %d: %d %03d\n", cas++, (int)(tt*100), quick_pow(n, k));
	} 
	return 0;
}

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
题意：给你一个数n，让你求这个数的k次方的前三位和最后三位

后三位可以用快速幂求出来。
前三位就要用到log了。先设10^p =n^k，同时取 log10，那么p=k* log10(n)。再设x=(int)p（整数部分），y=p-x（小数部分），那么10^p = 10^x * 10^ y；由于10^x 是10的倍数，那么10^ y=n^k /10^x，就接近 n^k 的值，只不过小数点在最高位后面，所以还要再乘100.

#include <iostream>
#include <cstdio>
#include <cmath>
#include <algorithm>
using namespace std;

int quick_pow(int n, int k){
	int ans=1;
	int base=n%1000;
	while (k){
		if (k&1)
			ans=ans*base%1000;
		base=base*base%1000;
		k>>=1;
	}
	return ans;
}
int main(){
	int n, k, t, cas=1;
	scanf("%d", &t);
	while (t--){
		scanf("%d%d", &n, &k);
		double p=(double)k*log10(n*1.0);
		p-=(int)p;
		double tt=pow(10.0, p);
		printf("Case %d: %d %03d\n", cas++, (int)(tt*100), quick_pow(n, k));
	} 
	return 0;
}

本文标签： 对数快速LightOJTrailingLeading