CodeForces 515C Drazil and Factorial (水题)

题意：给出含有 n 个只有阿拉伯数字的字符串a，设定函数F(a) = 每个数字的阶乘乘积 。需要找出 x，使得F(x) = F(a)，且组成 x 的数字中没有0和1。求最大的 x 为多少。

析：最大，那么首先是位数最多，然后是前面尽量大，所以我们要让位数最大，那么就转化，2-2， 3-3， 4-322， 5-5， 6-53， 7-7， 8-7222， 9-733.

然后排序就好了。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 5e4 + 5;
const LL mod = 10000000000007;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;

}
string str[] = {"", "", "2", "3", "322", "5", "53", "7", "7222", "7332"};
string s;

int main(){
    while(scanf("%d", &n) == 1){
        cin >> s;
        string ans = "";
        for(int i = 0; i < n; ++i)
            ans += str[s[i]-'0'];
        sort(ans.begin(), ans.end());
        for(int i = ans.size()-1; i >= 0; --i)
            printf("%c", ans[i]);
        printf("\n");
    }
    return 0;
}