lightoj1282：Leading and Trailing（数论）

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1282 - Leading and Trailing

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Time Limit: 2 second(s)

Memory Limit: 32 MB

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	Output for Sample Input
5 123456 1 123456 2 2 31 2 32 29 8751919	Case 1: 123 456 Case 2: 152 936 Case 3: 214 648 Case 4: 429 296 Case 5: 665 669

 

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PROBLEM SETTER: SHAMIM HAFIZ

SPECIAL THANKS: JANE ALAM JAN (SOLUTION, DATASET)

题意：求n^k前三位数和后三位数，若后者不足三位数补前导零。

思路：后三位用快速幂取模，前三位科学记数法表示为N^k = (double)a * 10^x，其中a为整数部分有3位数的浮点数，x为N^k位数减三，即x = (int)lg(n^k) - 2，随后移项两边取对数化简计算得到a，取a的整数部分即答案。

# include <stdio.h>
# include <math.h>

int fun(long long a, long long b)
{
    long long ret = 1, pow = a;
    while(b != 0)
    {
        if(b & 1) ret = (ret * pow) % 1000;
        pow = (pow * pow) % 1000;
        b >>= 1;
    }
    return (int)ret;
}

int main()
{
    int t, cas=1;
    long long n, k;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%lld%lld",&n,&k);
        double d = k*log10(n);
        d = d - (long long)d + 2;
        printf("Case %d: %lld %03d\n",cas++, (long long)pow(10, d), fun(n, k));

    }
    return 0;
}