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POJ1365_Prime Land【质因数分解】【素数】【水题】

2024-07-22 22:53:22   219人阅读

Prime Land
Time Limit: 1000MS Memory Limit: 10000K
Total Submissions: 3086Accepted: 1416
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 ekx, ekx-1, ..., e1, e0, (ekx > 0), that  The sequence 


(ekx, ekx-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


17 1
5 1 2 1
509 1 59 1
0
Sample Output


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


Central Europe 1997

题目大意：告诉你一个数的质因数x的所有底数pi和幂ei，输出x-1的质因数的所有底数和幂

思路：这道题不难，但是题意特别不好理解，对于我这种英文渣的人，愣是一个小时没看明白

关于题意举例说明吧  

例如 509 1 59 1  

x = 509^1 * 59^1 = 30031

x-1 = 30030

则答案13 1 11 1 7 1 5 1 3 1 2 1 就是 x-1 = 13^1 * 11^1 * 7^1 * 5^1 *3^1 *2^1 = 30031

那么直接按着题意暴力解决就行了。。。。。

#include<stdio.h>
#include<string.h>
#include<math.h>
/*
pow函数说明
原型：extern float pow(float x, float y);
用法：#include <math.h>
功能：计算x的y次幂。
说明：x应大于零，返回幂指数的结果。
*/
double p[110],e[110];
int Prime[35000],E[35000];

void IsPrime()
{
    Prime[0] = Prime[1] = 0;
    for(int i = 2; i <= 35000; i++)
    {
        Prime[i] = 1;
    }
    for(int i = 2; i <= 35000; i++)
    {
        for(int j = i+i; j <= 35000; j+=i)
        {
            Prime[j] = 0;
        }
    }
}
int main()
{
    int count,sign;
    IsPrime();
//    for(int i = 2; i <= 35000; i++)
//        if(Prime[i])
//            printf("%d ",i);
    while(1)
    {
        count = 0,sign = 0;
        memset(p,0,sizeof(p));
        memset(e,0,sizeof(e));
        memset(E,0,sizeof(E));

        while(1)
        {
            scanf("%lf",&p[count]);
            if(p[count] == 0)
            {
                sign = 1;
                 break;
            }
            scanf("%lf",&e[count]);
            count++;
            char c = getchar();
            if(c=='\n')
                break;
        }

        if(sign == 1)
            break;

        double num = 1;
        for(int i = 0; i < count; i++)
            num *= pow(p[i],e[i]);

        int sum = (int)num - 1;
//        printf("%d\n",sum);
        int flag = 0,pos = 2;
        for(int i = 2; i <= 32767; i++)
        {
            if(sum == 1)
                break;
            if(Prime[i])
            {
                while(sum % i == 0)
                {
                    E[i]++;
                    sum /= i;
                    if(flag == 0)
                    {
                        flag = 1;
                        pos = i;
                    }
                }
            }
        }

        for(int i = 32767; i>= 2; i--)
        {
            if(E[i]!=0 && i!=pos)
                printf("%d %d ",i,E[i]);
            else if(E[i]!=0 && i==pos)
            {
                printf("%d %d\n",i,E[i]);
                break;
            }
        }
    }

    return 0;
}



POJ1365_Prime Land【质因数分解】【素数】【水题】


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