Prime Land

链接：http://acm.hust.edu.cn/vjudge/problem/21459/origin

题目：Everybody in the Prime Land is using a prime base number system. In this system, each positive integer x is represented as follows: Let {pi}i=0,1,2,... denote the increasing sequence of all prime numbers. We know that x > 1 can be represented in only one way in the form of product of powers of prime factors. This implies that there is an integer kx and uniquely determined integers e _kx, e _kx-1, ..., e ₁, e ₀, (e _kx > 0), that The sequence

(e _kx, e _kx-1, ... ,e ₁, e ₀)



is considered to be the representation of x in prime base number system.

It is really true that all numerical calculations in prime base number system can seem to us a little bit unusual, or even hard. In fact, the children in Prime Land learn to add to subtract numbers several years. On the other hand, multiplication and division is very simple.

Recently, somebody has returned from a holiday in the Computer Land where small smart things called computers have been used. It has turned out that they could be used to make addition and subtraction in prime base number system much easier. It has been decided to make an experiment and let a computer to do the operation ``minus one''.

Help people in the Prime Land and write a corresponding program.

For practical reasons we will write here the prime base representation as a sequence of such pi and ei from the prime base representation above for which ei > 0. We will keep decreasing order with regard to pi.

题意：给出一个数x的所有质因数以及出现次数，求x-1的质因数及出现次数。

分析：挺吓人的一道题，，结果就是个稍加修饰的暴力题，，吧所给的数求出来（用快速幂优化），减一，分解质因数（记录出现次数），降序输出。。。然后就完了。。。

题解：

#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <algorithm>
#include <queue>
#include <stack>
#include <vector>
#include <map>
#include <string>
#include <cstring>
#include <functional>
#include <cmath>
#include <cctype>
#include <cfloat>
#include <climits>
#include <complex>
#include <deque>
#include <list>
#include <set>
#include <utility>
using namespace std;

__int64 mi(__int64 a,__int64 b)
{
	__int64 ans=1;
	__int64 temp=a;
	while(b){
		if(b&1)ans*=temp;
		b/=2;
		temp=temp*temp;
	}
	return ans;
}

int main()
{
	//freopen("in.txt","r",stdin);
	string s;
	while(getline(cin,s)){
		if(s=="0")
			return 0;
		stringstream ss(s);
		map<__int64,int> ma;
		__int64 a,b;
		__int64 sum=1;
		while(ss>>a>>b)
			sum*=mi(a,b);
		sum--;
		for(__int64 i=2;i*i<=sum;i++){
			while(sum%i==0){
				ma[i]++;
				sum/=i;
			}
		}
		if(sum!=1)
			ma[sum]++;
		map<__int64,int>::reverse_iterator it=ma.rbegin();
		for(;it!=ma.rend();it++){
			printf("%lld %d ",it->first,it->second);
		}
		printf("\n");
	}
	return 0;
}