POJ 1365 Prime Land_everybody in the prime land is using a prime base -优快云博客

本文链接：https://blog.youkuaiyun.com/u014737310/article/details/38350393

本文介绍了一种基于素数基数的特殊数值表示方法，并提供了一个程序实现，用于完成特定数值运算——即输入一个正整数的素数基数表示形式，输出该数减一后的素数基数表示。

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Description

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_{k_x}, e_{k_x-1}, ..., e₁, e₀, (e_{k_x} > 0), that

The sequence

(e_{k_x}, e_{k_x-1}, ... ,e₁, e₀)

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.

Input

The input consists of lines (at least one) each of which except the last contains prime base representation of just one positive integer greater than 2 and less or equal 32767. All numbers in the line are separated by one space. The last line contains number 0.

Output

The output contains one line for each but the last line of the input. If x is a positive integer contained in a line of the input, the line in the output will contain x - 1 in prime base representation. All numbers in the line are separated by one space. There is no line in the output corresponding to the last ``null'' line of the input.

Sample Input

Sample Output

2 4
3 2
13 1 11 1 7 1 5 1 3 1 2 1

Source

题目大意，输入一行数，例如 a,b,c,d,e,f...... 然后问 a^b*c^d*e^f*.... 的结果减去1后由哪些素数可以乘得，输出素数（由大到小）和个数，

对于此题我只想说 cmath 的pow（快速幂）函数对整型有误差，且用且珍惜，最好手写，(⊙o⊙)…

还有一点，题目说不会超 int 所以不用担心什么了

上代码了，解释在代码中

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;

#define N 100005

int a[N],b[N];

struct stud{
int a,len;
}f[1000];

int k=0;


void inset()   //素数表
{
    int i,j;
    for(i=2;i<N;i++)
        if(a[i]==0)
    {
        b[k++]=i;
        for(j=i*2;j<N;j+=i)
            a[j]=1;
    }
}

int pow(int a,int b)  // a^b  快速幂函数
{
    int n;
    n=1;
    while (b)
    {
        if (b%2==1)
            n=n*a;
        a=a*a;
        b=b/2;
    }
    return n;
}

int main()
{
    int i,j;
    int n,m,sum;

    inset();

    while(scanf("%d",&n),n)
    {
        scanf("%d",&m);
        sum=pow(n,m);

        char c;
        while(scanf("%c",&c)&&c!='\n')  //非常巧妙的地方哟
        {
            scanf("%d%d",&n,&m);
            sum*=pow(n,m);
        }
        sum--;

       int te=0;

       for(i=0;i<k&&sum!=1;i++)
        if(sum%b[i]==0)
       {
           f[te].a=b[i];      //保存该素数
           f[te].len=0;

           while(sum%b[i]==0)
           {
               f[te].len++;   //记录该素数个数
               sum/=b[i];    
           }
           te++;
       }

       for(i=te-1;i>=0;i--)
           if(i==0)
              printf("%d %d\n",f[i].a,f[i].len);
       else
              printf("%d %d ",f[i].a,f[i].len);

    }
    return 0;
}