Leading and Trailing [数学]

最新推荐文章于 2025-09-04 12:25:10 发布

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本文介绍了一种高效算法来找出两个整数n和k相乘后的幂nk的最显著的三个高位数字和三个低位数字。通过使用对数和模运算，此算法能够在不直接计算完整幂值的情况下得出答案。

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

题解

解题方法写在这儿

代码如下

#include<stdio.h>
#include<cmath>
typedef long long LL;

LL _pow(LL a,int n,LL MOD){
    LL ans=1;
    while(n){
        if(n%2) ans=ans*a%MOD;
        a=a*a%MOD;
        n>>=1;
    }
    return ans;
}

int main()
{
    LL n;int T,k;
    scanf("%d",&T);
    for(int t=1;t<=T;t++){
        scanf("%lld%d",&n,&k);
        double a=(double)k*log10(n)-floor((double)k*log10(n));
        a=pow(10.0,a);
        a=floor(100*a);
        LL b=_pow(n,k,1000);
        printf("Case %d: %0.0lf %0.3lld\n",t,a,b);
    }
    return 0;
}