Leading and Trailing - UVa 11029

Apart from the novice programmers, all others know that you can’t exactly represent numbers raised to some high power. For example, the C function pow(125456, 455) can be represented in double data type format, but you won’t get all the digits of the result. However we can get at least some satisfaction if we could know few of the leading and trailing digits. This is the requirement of this problem.

 

Input

 

The first line of input will be an integer T<1001, where T represents the number of test cases. Each of the next T lines contains two positive integers, n and k. n will fit in 32 bit integer and k will be less than 10000001.

 

Output

 

For each line of input there will be one line of output. It will be of the format LLL…TTT, where LLL represents the first three digits ofn^k and TTT represents the last three digits of n^k. You are assured that n^k will contain at least 6 digits.

 

 

Sample Input

Output for Sample Input

2

123456 1

123456 2

123...456

152...936

题意：求出n^k的前三位和后三位。

思路：后三位用快速幂求，前三位n^k=pow(10,k*log(n))，再保留三位取整。

AC代码如下：

#include<cstdio>#include<cstring>#include<cmath>using namespace std;int MOD=1000;int main(){    int T,t,n,m,i,j,k,f,ret,p,q,h;    scanf("%d",&T);    for(t=1;t<=T;t++)    {        scanf("%d%d",&n,&m);        ret=1;f=n%MOD;        k=m;        while(k)        {            if(k&1)              ret=ret*f%MOD;            f=f*f%MOD;            k/=2;        }        q=floor(pow(10,2+fmod(m*log10(n),1) ));        printf("%d...%03d\n",q,ret);    }}