2016-12-02 12:52:57来源:网络收集作者:路过秋天人点击

Description
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

d可以由fmod(a*k,1)求出，fmod的作用是求浮点数的模，所以这个式子能求出a*k小于1的部分即小数部分。前面又可以根据n=10^a得a=log10(n)。

#include
#include
#include
#include
#include
#include
#include
#include
#include
#define INF 0x3f3f3f3f
#define MAXN 1000005
#define mod 1000000007
using namespace std;
int pow_mod(int a, int n,int m)
{
if(n==1)
return a%m;
int x=pow_mod(a,n/2,m);
long long ans=(long long)x*x%m;
if(n&1)
ans=ans*a%m;
return (int)ans;
}
int main()
{
int t,cnt=1;
long long n,k;
scanf("%d",&t);
while(t--)
{
scanf("%lld%lld",&n,&k);
int trail=pow_mod(n,k,1000);
}
return 0;
}

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