codeforces 515c Drazil and Factorial 分类讨论
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题意:
略
思路:
每个阶乘肯定是分得越多个阶乘数就越好。
分类讨论。
1. 0和1直接忽略。
2. { 2,3,5,7 }这些数的阶乘是不能再分得,原数即可。
3. 4:{3, 2, 2},6:{5, 3} 8:{7, 2, 2, 2} 9:{7, 3, 3, 2}。
code:
#include <bits/stdc++.h>using namespace std;int n;string str;int prime[] = {2, 3, 5, 7};int t[] = {4, 6, 8, 9};int a[][5] = {{3, 2, 2}, {5, 3}, {7, 2, 2, 2}, {7, 3, 3, 2}};int cnt[10];void solve() { for(int i = 0;i < n; i++) { if(str[i] == '1'|| str[i] == '0') continue; bool flag = false; for(int j = 0;j < 4; j++) { if(str[i] == prime[j]+'0') { cnt[prime[j]]++; flag = true; break; } } if(flag) continue; for(int j = 0;j < 4; j++) { if(str[i] == t[j]+'0') { for(int z = 0;z < 5; z++) { cnt[a[j][z]]++; } break; } } } for(int i = 9; i >= 1; i--) { for(int j = 0;j < cnt[i]; j++) cout<<i; } cout<<endl;} int main() { cin>>n>>str; solve(); return 0;}
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