Prime Land

http://poj.org/problem?id=1365

题意：给定一个数字n的拆分形式，然后让你求解n-1的值；

解析：直接爆搞

// File Name: poj1365.cpp
// Author: bo_jwolf
// Created Time: 2013骞?0鏈?9鏃?鏄熸湡涓?21:29:25

#include<vector>
#include<list>
#include<map>
#include<set>
#include<deque>
#include<stack>
#include<bitset>
#include<algorithm>
#include<functional>
#include<numeric>
#include<utility>
#include<sstream>
#include<iostream>
#include<iomanip>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<cstring>
#include<ctime>

using namespace std;
vector<int> prime, ans;
const int maxn = 40000;
int unprime[ maxn ];
int main(){
	unprime[ 0 ] = unprime[ 1 ] = true;
	for( int i = 2; i < maxn; ++i ){
		if( !unprime[ i ] ){
			prime.push_back( i );
			for( int j = i + i; j < maxn; j += i ){
				unprime[ j ] = true;
			}
		}
	}
	int n, p;
	string line;
	while( getline( cin, line ),line[ 0 ] != 48 ){
		ans.clear();
		istringstream stream( line );
			long long sum = 1;
		while( stream >> n >> p ){
			while( p-- ){
				sum *= n;
			}
		}		sum -= 1;
		for( int i = prime.size() - 1; i >= 0; --i ){
			if( sum % prime[ i ] == 0 ){
							ans.push_back( prime[ i ] );
				int temp = 0;
				while( sum % prime[ i ] == 0 ){
					sum /= prime[ i ];
					temp++;
				}
				ans.push_back( temp );
			}
		}
		for( int i = 0 ; i < ans.size(); ++i ){
			cout << ans[ i ] << ( i == ans.size() - 1 ?'\n':' ' );
		}
	}
return 0;
}