Drazil is playing a math game with Varda.
Let’s define for positive integer x as a product of factorials of its digits. For example, .
First, they choose a decimal number a consisting of n digits that contains at least one digit larger than 1. This number may possibly start with leading zeroes. Then they should find maximum positive number x satisfying following two conditions:
x doesn’t contain neither digit 0 nor digit 1.
= .
Help friends find such number. Input
The first line contains an integer n (1≤n≤15) — the number of digits in a.
The second line contains n digits of a. There is at least one digit in a that is larger than 1. Number a may possibly contain leading zeroes. Output
Output a maximum possible integer satisfying the conditions above. There should be no zeroes and ones in this number decimal representation. Sample test(s) Input
4 1234
Output
33222
Input
3 555
Output
555
Note
In the first case,
水题, 将阶乘分解成素数的阶乘即可
/*************************************************************************> File Name: cf292-A.cpp> Author: ALex> Mail: zchao1995@gmail.com> Created Time: 2015年02月18日 星期三 00时26分55秒 ************************************************************************/;const double pi = acos(-1.0);const int inf = 0x3f3f3f3f;const double eps = 1e-15;LL;typedef pair <int, int> PLL;char str[20];int num[10];int main (){int n;while (~scanf(“%d”, &n)){scanf(“%s”, str);memset (num, 0, sizeof(num));int len = strlen (str);for (int i = 0; i < len; ++i){if (str[i] == ‘0’ || str[i] == ‘1’){continue;}else if (str[i] == ‘2’ || str[i] == ‘3’ || str[i] == ‘5’ || str[i] == ‘7’){num[str[i] – ‘0’]++;}else if (str[i] == ‘4’){num[2] += 2;num[3]++;}else if (str[i] == ‘6’){num[5]++;num[3]++;}else if (str[i] == ‘8’){num[7]++;num[2] += 3;}else{num[7]++;num[3] += 2;num[2]++;}}for (int i = 9; i >= 2; –i){for (int j = 1; j <= num[i]; ++j){printf(“%d”, i);}}printf(“\n”);}return 0;}
，玩坏了可以选择重来，
