Codeforces 515C 题解(贪心+数论)(思维题)

题面

传送门:http://codeforces.com/problemset/problem/515/C

Drazil is playing a math game with Varda.

Let’s define f(x) f ( x ) for positive integer x as a product of factorials of its digits. For example, f(135)=1!∗3!∗5! f ( 135 ) = 1 ! ∗ 3 ! ∗ 5 !

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:

1. x doesn’t contain neither digit 0 nor digit 1.

2. f(x)=f(a) f ( x ) = f ( a )

Help friends find such number.

题目大意:
定义f(x)为x的每一位的阶乘之积
给出一个数a,求满足条件(x的每一位没有0或1)的最大x,使f(x)=f(a)

分析:

此题可用贪心求解
贪心的思路是很显然的,应该让x的位数尽量多,而每一位尽量小,最大的数应该排在最左边
这样,我们就可以把a的每一位拆开
如a=6
6!=6*5*4*3*2*1=6*5!=3!*5!
所以6可以被替换成35

所以我们把0~9的数字拆开(其实0,1应该直接舍去,因为不符合条件)
0!=0!
1!=1!
2!=2!
3!=3!
4!=3!*4=3!*2!*2!
5!=5!
6!=5!*6=5!*3!
7!=7!
8!=7!*8=7!*2!*2!*2!
9!=9*8*7!=7!*3!*3!*2!;

代码:

#include
#include
#include
#include
using namespace std;
const string convert[10]={"0","0","2","3","322","5","53","7","7222","7332"};
string s;
string ans;
int n;
int cmp(char x,char y){
    return x>y;
}
int main(){
    scanf("%d",&n);
    cin>>s;
    ans="";
    for(int i=0;iif(s[i]-'0'==0||s[i]-'0'==1) continue;
        ans=ans+convert[s[i]-'0'];
    }
//  cout<
    sort(ans.begin(),ans.end(),cmp);
    cout<