https://codeforces.com/contest/1176/problem/A

題意：采取三種操作，使得n變成1的步驟數；否則-1

題解：模擬

1、n分解成質數乘積；

2、ans=2的個數+3的個數*2+5的個數*3；

/* *@Author: STZG *@Language: C++ */ #include <bits/stdc++.h> #include<iostream> #include<algorithm> #include<cstdlib> #include<cstring> #include<cstdio> #include<string> #include<vector> #include<bitset> #include<queue> #include<deque> #include<stack> #include<cmath> #include<list> #include<map> #include<set> //#define DEBUG #define RI register int #define endl "\n" using namespace std; typedef long long ll; //typedef __int128 lll; const int N=100000+10; const int M=100000+10; const int MOD=1e9+7; const double PI = acos(-1.0); const double EXP = 1E-8; const int INF = 0x3f3f3f3f; ll t,n,m,k,p,l,r,u,v; int ans,cnt,flag,temp,sum; int a[N]; char str; struct node{}; int main() { #ifdef DEBUGfreopen("input.in", "r", stdin);//freopen("output.out", "w", stdout); #endif//ios::sync_with_stdio(false);//cin.tie(0);//cout.tie(0);scanf("%I64d",&t);while(t--){scanf("%I64d",&n);ans=0;while(n%2==0)n=n/2,ans++;while(n%3==0)n=n/3,ans+=2;while(n%5==0)n=n/5,ans+=3;cout<<(n==1?ans:-1)<<endl;}#ifdef DEBUGprintf("Time cost : %lf s\n",(double)clock()/CLOCKS_PER_SEC); #endif//cout << "Hello world!" << endl;return 0; }

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