新生42

数字反转

数字反转：

题目描述

给定一个整数，请将该数各个位上数字反转得到一个新数。新数也应满足整数的常见形式，即除非给定的原数为零，否则反转后得到的新数的最高位数字不应为零。

输入

输入共 1 行，一个整数N。-1,000,000,000 ≤ N≤ 1,000,000,000

输出

输出共 1 行，一个整数，表示反转后的新数。

样例输入 Copy

123

样例输出 Copy

321
不想说啥了。写的挺长...

#include<iostream>
#include<algorithm>
using namespace std;
int main(){
    string ch;
    cin>>ch;
    if(ch[0]=='-')
        reverse(ch.begin()+1,ch.end());
    else
        reverse(ch.begin(),ch.end());
    if(ch[0]=='-')
    {
        if(ch[1]!='0')
            cout<<ch;
        else
        {
            int i=1;
            while(ch[i]=='0')
                i++;
            cout<<"-";
            for(int j=i;j<ch.size();j++)
                cout<<ch[j];
        }
    }
    else
    {
        if(ch[0]!='0')
            cout<<ch;
        else
        {
            int i=0;
            while(ch[i]=='0')
                i++;
            for(int j=i;j<ch.size();j++)
                cout<<ch[j];
                }
    }
}

题目描述

The word internationalization is sometimes abbreviated to i18n. This comes from the fact that there are 18 letters between the first i and the last n.
You are given a string s of length at least 3 consisting of lowercase English letters. Abbreviate s in the same way.

Constraints
3≤|s|≤100 (|s| denotes the length of s.)
s consists of lowercase English letters.

 

输入

Input is given from Standard Input in the following format:
s

输出

Print the abbreviation of s.

样例输入 Copy

internationalization

样例输出 Copy

i18n
这个题nnd，我的问题吗，我一开始一直以为是第一个i和最后一个n之间有多少个数字，每一次修改最后答案都只对17，后来才发现....第一个数和最后一个数之间有多少个单词，再输出开头和结尾就行了...
没啥好说的了

We have a sequence of length N, a=(a1,a2,…,aN). Each ai is a positive integer.
Snuke's objective is to permute the element in a so that the following condition is satisfied:
For each 1≤i≤N−1, the product of ai and ai+1 is a multiple of 4.
Determine whether Snuke can achieve his objective.

Constraints
2≤N≤105
ai is an integer.
1≤ai≤109

输入

Input is given from Standard Input in the following format:
N
a1 a2 … aN

输出

If Snuke can achieve his objective, print Yes; otherwise, print No.

样例输入 Copy

3
1 10 100

样例输出 Copy

Yes

提示

One solution is (1,100,10).

意思就是重新排列使相邻的两个数相乘可以被4整除

ac85的代码：....

#include<iostream>
using namespace std;
int main(){
    int n,t,j=0,o=0,p=0,x=0;
    cin>>n;
    for(int i=0;i<n;i++)
    {
        cin>>t;
        if(t%2==1)
            j++;
        else if(t%4==0)
            p++;//既可以被4整除又一定是偶数
        else if(t%2==0&&t%4!=0)
            x++;
    }
    if(j==1)
    {
        if(p>=1)
            cout<<"Yes"<<endl;//可以放两端 
        else
            cout<<"No"<<endl;
    }
    else if(j==2)//两端 
    {
        if(p==1&&x>=2||p>1)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    } 
    else if(j>2)//中间
    {
        j-=2;
        if(p>=j+1&&x%2==0)
            cout<<"Yes"<<endl;
        else
            cout<<"No"<<endl;
    }
    else if(j==0)
        cout<<"Yes"<<endl;
}

到后面越写越乱....