Leading and Trailing

最新推荐文章于 2020-01-21 21:40:09 发布

转载最新推荐文章于 2020-01-21 21:40:09 发布 · 118 阅读

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原文链接：http://www.cnblogs.com/joeylee97/p/6831310.html

本文介绍了一种高效算法，用于求解大整数n的k次幂的最前三位和最后三位数字。通过快速幂取模运算获取最后三位，利用对数小数部分计算前三位。附带C++实现代码。

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

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 n^k contains at least six digits.

Sample Input

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

后三位数字可以通过快速幂取模运算来获得，前三位数字可以通过对数的小数部分来获得！

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstring>
#include<sstream>
#include<algorithm>
#include<queue>
#include<vector>
#include<cmath>
#include<map>
#include<stack>
#include<set>
#include<fstream>
#include<memory>
#include<list>
#include<string>
using namespace std;
typedef long long LL;
typedef unsigned long long ULL;
#define MAXN  1000010
#define LLL 1000000000
#define INF 1000000009
LL Pow(LL x,LL k)
{
    LL ret = 1;
    while (k)
    {
        if (k & 1!=0)
            ret = (ret*x)%1000;
        x = x*x;
        x %= 1000;
        k /= 2;
    }
    return ret;
}
int main()
{
    int T;
    LL n, k;
    cin >> T;
    for (int cas = 1; cas <= T; cas++)
    {
        scanf("%lld%lld", &n, &k);
        LL tmp = n % 1000;
        LL ans2 = Pow(tmp, k)%1000,ans1;
        double num = k*log10(n*1.0);
        num -= LL(num);
        ans1 = LL(pow(10, num) * 100);
        printf("Case %d: %lld %03lld\n", cas, ans1, ans2);
    }
    return 0;
}

转载于:https://www.cnblogs.com/joeylee97/p/6831310.html