poj 2720 Last Digits

Last Digits

Time Limit: 1000MS		Memory Limit: 65536K
Total Submissions: 2233		Accepted: 474

Description

Exponentiation of one integer by another often produces very large results. In this problem, we will compute a function based on repeated exponentiation, but output only the last n digits of the result. Doing this efficiently requires careful thought about how to avoid computing the full answer.

Given integers b, n, and i, we define the function f(x) recursively by f(x) = b^f(x-1) if x > 0, and f(0)=1. Your job is to efficiently compute the last n decimal digits of f(i). 

Input

The input consists of a number of test cases. Each test case starts with the integer b (1 <= b <= 100) called the base. On the next line is the integer i (1 <= i <= 100) called the iteration count. And finally, the last line contains the number n (1 <= n <= 7), which is the number of decimal digits to output. The input is terminated when b = 0.

Output

For each test case, print on one line the last n digits of f(i) for the base b specified. If the result has fewer than n digits, pad the result with zeroes on the left so that there are exactly n digits.

Sample Input

2
4
7
10
10
6
3
10
7
0

Sample Output

0065536
000000
4195387

Source

Rocky Mountain 2005

/*
* @Author: Lyucheng
* @Date:   2017-08-07 15:47:29
* @Last Modified by:   Lyucheng
* @Last Modified time: 2017-08-07 20:10:43
*/
/*
 题意：定义一个函数f(i)=b^(f(i-1)),给你b,i,n让你求f(i)的后n位，不足的用前导零补充

 思路：后n位就是f(n)%(10^n)

 问题：超时...打表
LL b,n,i;
LL mod;
char str[10];
LL pos;
char format[] = "%00d\n";

inline LL power(LL a,LL b,LL mod){//a的b次方
    if(b==0) return 1;
    LL cnt=power(a,b/2,mod);
    cnt=cnt*cnt%mod;
    if(b%2==1) cnt=cnt*a%mod;
    return cnt;
}

// return f(x)%mod
inline LL fun(LL b,LL x,LL mod){
    if(x==0) return 1LL;//如果是0次，那么就是1
    else{
        LL res=power(b,fun(b,x-1,mod),mod);
        if(res==0) res=mod;
        return res;
    }
}
*/
#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <vector>
#include <queue>
#include <set>
#include <map>
#include <string>
#include <math.h>
#include <stdlib.h>
#include <time.h>

using namespace std;

int a[][]={
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},
{,,,,,,,,,,,,},

};
int b,i,n;

int main(){
    // freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
    while(scanf("%d",&b)!=EOF&&b){
        scanf("%d%d",&i,&n);
        string s="";
        int pos=a[b-][(i>?:i)];
        for(int i=;i<n;i++){
            s+=pos%+'';
            pos/=;
        }
        while(n--){
            cout<<s[n];
        }cout<<endl;
    }
    return ;
}