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最新推荐文章于 2020-10-04 10:29:23 发布

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最新推荐文章于 2020-10-04 10:29:23 发布

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分类专栏：数论

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本文介绍了一种求解n的k次方的最前三位和最后三位数字的方法。通过使用快速幂运算来获取最后三位，并利用对数将数值范围缩小以计算最前三位。

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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

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

题意就是求n的k次方的前三位和后三位，后三位你用快速幂就可以搞定，前三位就比较麻烦了，开始我用的while循环去弄，结果超时，后来改成字符串也弄不了。然后看了大佬的博客才明白，原来利用对数就可以把数的范围缩小，然后再用while就可以了。

#include<stdio.h>
#include<math.h>
long long int cn(long long int x, int y)
{
long long int z=1;
while(y)
{
if(y%2)
z=z*x%1000;
x=x*x%1000;
y=y/2;
}
return z;
}
int main()
{
int t;
scanf("%d",&t);
for(int i=1; i<=t; i++)
{
int n,k;
scanf("%d%d",&n,&k);
int l=k*log10(n);
double t=pow(10,1.0*k*log10(n)-l);
while(t<100)
t*=10;
int ans=t;
int sum=cn(n,k);
printf("Case %d: %d %03d\n",i,ans,sum);
}
return 0;
}