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

5530

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;}