UVa 160 Factors and Factorials

Factors and Factorials

The factorial of a number N (written N!) is defined as the product of all the integers from 1 to N. It is often defined recursively as follows:

Factorials grow very rapidly--5! = 120, 10! = 3,628,800. One way of specifying such large numbers is by specifying the number of times each prime number occurs in it, thus 825 could be specified as (0 1 2 0 1) meaning no twos, 1 three, 2 fives, no sevens and 1 eleven.

Write a program that will read in a number N (  ) and write out its factorial in terms of the numbers of the primes it contains.

Input

Input will consist of a series of lines, each line containing a single integer N. The file will be terminated by a line consisting of a single 0.

Output

Output will consist of a series of blocks of lines, one block for each line of the input. Each block will start with the number N, right justified in a field of width 3, and the characters `!', space, and `='. This will be followed by a list of the number of times each prime number occurs in N!.

These should be right justified in fields of width 3 and each line (except the last of a block, which may be shorter) should contain fifteen numbers. Any lines after the first should be indented. Follow the layout of the example shown below exactly.

Sample input

5
53
0

Sample output

  5! =  3  1  1
 53! = 49 23 12  8  4  4  3  2  2  1  1  1  1  1  1
        1

思路：可以先做一张素数表，不用太多，因为题目要求是输入2~100的数，所以我们找出100以内的素数即可。接下来对每次输入都不断的用素数表中每一个数去除它，同一个数要除到不能再除了为止，最后注意打印的格式即可。格式务必注意当输出恰好为15个数的时候是不输出那个换行和那些空格的。

AC代码（0.006s）：

#include <stdio.h>

int IsPrime(int n)
{
    int i;
    for (i = 2; i * i <= n; i++) {
        if (n % i == 0) {
            return 0;
        }
    }
    return 1;
}

int main(int argc, const char * argv[]) {
    
    int i, count = 0;
    int prime[100] = {0};
    for (i = 2; i < 100; i++) {
        if (IsPrime(i)) {
            prime[count++] = i;
        }
    }
    
    int input, j;
    while (scanf("%d", &input)) {
        if (input == 0) {
            break;
        }
        
        int ans[100] = {0}, maxCur = -1;
        for (i = 2; i <= input; i++) {
            int m = i;
            for (j = 0; j < count; j++) {
                
                if (m < prime[j]) {
                    break;
                }
                
                while (m % prime[j] == 0) {
                    ans[j]++;
                    m /= prime[j];
                    if (j > maxCur) {
                        maxCur = j;
                    }
                }
            }
            
        }
        
        int numCount = 0;
        printf("%3d! =", input);
        for (i = 0; i <= maxCur; i++) {
            printf("%3d", ans[i]);
            numCount++;
            if (numCount == 15) {
                numCount = 0;
                if (i + 1 <= maxCur) {
                    printf("\n      ");
                }
            }
        }
        printf("\n");
    }
    
    return 0;
}