ZQUOJ1403Factoring Large Numbers解题报告

 Factoring Large Numbers

memory limit: 65536KB time limit: 500MS

accept: 17 submit: 48

Description

One of the central idea behind much cryptography is that factoring large numbers is computationally intensive. In this context one might use a 100 digit number that was a product of two 50 digit prime numbers. Even with the fastest projected computers this factorization will take hundreds of years.

You don't have those computers available, but if you are clever you can still factor fairly large numbers.

Input

The input will be a sequence of integer values, one per line, terminated by a negative number. The numbers will fit in gcc's long long int datatype.

Output

Each positive number from the input must be factored and all factors (other than 1) printed out. The factors must be printed in ascending order with 4 leading spaces preceding a left justified number, and followed by a single blank line.

Sample Input

90
1234567891
18991325453139
12745267386521023
-1

Sample Output

2
3
3
5

1234567891

3
3
13
179
271
1381
2423

30971
411522630413

Source

ZQUCPC个人赛5

Author

Waterloo Local

 

题意：

对输入的数分解质因数。需要用到64位整型。

 

问题的分析与解题思路：

分解质因数，一开始的想法都很容易会对该数的因数进行素数判断，其实不必。因为题意输出是从小到大的，因此只要从i=2开始分解，直到没办法被i=2分解时，i自增。这样不必对i进行素数判断，因为所有不是素数的数（设为a）肯定有它自己的质因数，而a的质因数肯定比a自身小，因此还没除到a时，a已经被分解了，以此类推，可以保证输出的所有数都是素数。

 

AC方法和数据结构：

直接从i=2开始对n进行分解，当i能被n整除时（即n%i==0），执行n/i，否则，i自增。当i*i大于n时，结束循环。

AC代码：

#include<stdio.h> int main() { __int64 n,i; //n为要分解的数，i为自变量 while(scanf("%I64d",&n)!=EOF) { if(n==-1) break; for(i=2;n>1;) //i从2开始试除 { if(n%i==0) //如果n能被i整除，说明i是n的一个质因数 { printf(" %I64d/n",i); //输出i之前注意要空四格 n/=i; } else { i++; if(i*i>n) { printf(" %I64d/n",n); //如果i*i>n时，说明此时的n是一个质因数，但要保证n不等于1 break; } } } printf("/n"); } return 0; }

总结：

注意技巧，不然容易超时。