Uva 10879 Code Refactoring

Problem B
Code Refactoring
Time Limit: 2 seconds

 

"Harry, my dream is a code waiting to be
broken. Break the code, solve the crime."

Agent Cooper

 

Several algorithms in modern cryptography are based on the fact that factoring large numbers is difficult. Alicia and Bobby know this, so they have decided to design their own encryption scheme based on factoring. Their algorithm depends on a secret code, K, that Alicia sends to Bobby before sending him an encrypted message. After listening carefully to Alicia's description, Yvette says, "But if I can intercept K and factor it into two positive integers, A and B, I would break your encryption scheme! And the K values you use are at most 10,000,000. Hey, this is so easy; I can even factor it twice, into two different pairs of integers!"

Input
The first line of input gives the number of cases, N (at most 25000). N test cases follow. Each one contains the code, K, on a line by itself.

Output
For each test case, output one line containing "Case #x: K = A * B = C * D", where A, B, C and D are different positive integers larger than 1. A solution will always exist.

 

Sample Input	Sample Output
3 120 210 10000000	Case #1: 120 = 12 * 10 = 6 * 20 Case #2: 210 = 7 * 30 = 70 * 3 Case #3: 10000000 = 10 * 1000000 = 100 * 100000

 

---

Problemsetter: Igor Naverniouk

 

#include<stdio.h>
#include<string.h>
#define MAXN 10000
int prime[MAXN];
int flag[MAXN];

is_prime()
{
    int i, j, n = 0;
    memset(flag, 0, sizeof(flag));
    for(i=2; i<MAXN; ++i)
    {
        if(!flag[i])
        {
            prime[n++] = i;
            flag[i] = 1;
            for(j=i+i; j<MAXN; j+=i)
            flag[j] = 1;
        }
    }
    
    return n;
}

int main()
{
    int term, track, i, T, cnt = 0, n, temp, a, b, c, d;
    term = is_prime();
    scanf("%d", &T);
    while(++cnt <= T)
    {
        scanf("%d", &n);
        temp = n;

        for(track=i=0; n != 1 && n != 0 && i<term; ++i)
        {
            while(n%prime[i] == 0) 
            {
                flag[track++] = prime[i];
                n /= prime[i];
            }
        }
        if(n != 1) flag[track++] = n;
        
        a = temp/flag[0], b = temp/a;
        for(c=d=1,i=0; i<track; ++i)
        {
            c *= flag[i];
            if(c != a && c != b) break;
        }
            
        d = temp/c;
        printf("Case #%d: %d = %d * %d = %d * %d\n", cnt, temp, a, b, c, d);
    }
    
    return 0;
}

解题思路：

这题比较简单，因为是特判题(Special judge)题目的要求很宽，题目的背景设得太渣了，什么  After listening carefully to Alicia's description 然后就没了逻辑性了，还是直接看Sample好，

思路跟上面几题一样，还是找因子，然后累乘找到能整除n的两个不同的数a, c ( a*c != n)，然后按题目要求输出就行了