LightOJ 1282 - Leading and Trailing （求n^k的前三位和后三位）

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 taskis to find the most significant three digits, and least significant threedigits 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 leadingdigits (most significant) and three trailing digits (least significant). Youcan assume that the input is given such that n^k contains atleast 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

 

思路：后三位很好求，快速幂取模就好了，同样前三位我们也可以模拟快速幂来写，用double存储答案，模拟一下快速幂，中间记得转换就好了，注意输出为3位，如果后三位是022，那么必须输出022，不能输出22

ac代码：

#include<stdio.h>
#include<math.h>
#include<string.h>
#include<stack>
#include<set>
#include<queue>
#include<vector>
#include<iostream>
#include<algorithm>
#define MAXN 101000
#define LL long long
#define ll __int64
#define INF 0xfffffff
#define mem(x) memset(x,0,sizeof(x))
#define PI acos(-1)
#define eps 1e-8
using namespace std;
ll gcd(ll a,ll b){return b?gcd(b,a%b):a;}
ll lcm(ll a,ll b){return a/gcd(a,b)*b;}
ll powmod(ll a,ll b,ll MOD){ll ans=1;while(b){if(b%2)ans=ans*a%MOD;a=a*a%MOD;b/=2;}return ans;}
double dpow(double a,ll b){double ans=1.0;while(b){if(b%2)ans=ans*a;a=a*a;b/=2;}return ans;}
//head
double change(double x)
{
    while(x>=1000.0)
    {
        x/=10.0;
    }
    return x;
}
int main()
{
    int t,i;
    int cas=0;
    scanf("%d",&t);
    while(t--)
    {
        ll n,k;
        scanf("%I64d%I64d",&n,&k);
        int ans2=(int)powmod(n,k,1000);
        double ans1=1.0,a=n*1.0;
        while(k)
        {
            if(k%2)
            {
                ans1=ans1*a;
                ans1=change(ans1);
            }
            a=a*a;
            a=change(a);
            k/=2;
        }
        printf("Case %d: %d %03d\n",++cas,(int)ans1,ans2);
    }
    return 0;
}