zoukankan      html  css  js  c++  java
  • Codeforces1176A(A题)Divide it!

    Divide it!

    You are given an integer nn.

    You can perform any of the following operations with this number an arbitrary (possibly, zero) number of times:

    1. Replace nn with n2n2 if nn is divisible by 22;
    2. Replace nn with 2n32n3 if nn is divisible by 33;
    3. Replace nn with 4n54n5 if nn is divisible by 55.

    For example, you can replace 3030 with 1515 using the first operation, with 2020 using the second operation or with 2424 using the third operation.

    Your task is to find the minimum number of moves required to obtain 11 from nn or say that it is impossible to do it.

    You have to answer qq independent queries.

    Input

    The first line of the input contains one integer qq (1q10001≤q≤1000) — the number of queries.

    The next qq lines contain the queries. For each query you are given the integer number nn (1n10181≤n≤1018).

    Output

    Print the answer for each query on a new line. If it is impossible to obtain 11 from nn, print -1. Otherwise, print the minimum number of moves required to do it.

    代码:

     1 #include<iostream>
     2 #include<algorithm>
     3 using namespace std;
     4 int main() {
     5     int q;
     6     long long a[1005];
     7     cin>>q;
     8     for(int i=0; i<q; i++) {
     9         cin>>a[i];
    10     }
    11     for(int q1=0;q1<q;q1++){
    12         int cnt=0;
    13         while(a[q1]!=1) {
    14             if(a[q1]%2!=0&&a[q1]%3!=0&&a[q1]%5!=0){
    15                 cnt=-1;
    16                 break;
    17             }
    18             if(a[q1]%2==0) {
    19                 a[q1]/=2;
    20                 cnt++;
    21             }
    22             if(a[q1]%3==0) {
    23                 a[q1]=a[q1]*2/3;
    24                 cnt++;
    25             }
    26             if(a[q1]%5==0) {
    27                 a[q1]=a[q1]*4/5;
    28                 cnt++;
    29             }
    30             if(a[q1]==0){
    31                 break;
    32             }
    33         }
    34         cout<<cnt<<endl;
    35     }
    36 }

    思路分析:数要是不能被3,5,2整除,就设置cnt为-1并输出。先除2/n再2n/3z再4n/5统计次数,最后化为1,输出cnt。

    链接:https://codeforces.com/problemset/problem/1176/A

  • 相关阅读:
    textarea中的空格与换行
    js判断微信内置浏览器
    关于express4不再支持body-parser
    html5 geolocation API
    屏幕密度与分辨率
    nodebeginer
    手机浏览器下IScroll中click事件
    iphone手机上的click和touch
    AngularJS学习笔记一
    不用bootstrap实现居中适应
  • 原文地址:https://www.cnblogs.com/yuanhang110/p/11267893.html
Copyright © 2011-2022 走看看