嘻嘻，简洁的代码来喽！！

#include<bits/stdc++.h>
using namespace std;
int main(){
    int n;
    cin >> n;
    if(n == 1) cout << "swim";
    else if(n == 3) cout << "program";
    else if(n == 5) cout << "read";
    else if(n == 6) cout << "math";
    else cout << "rest";
    return 0;
}

暴 力 解 法``

#include<iostream>
using namespace std;
int main()
{
    int a;
    cin>>a;
    if (a==3)
        cout<<"program";
    if (a==1)
        cout<<"swim";
    if (a==5)
        cout<<"read";
    if (a==6)
        cout<<"math";
    if (a==2)
        cout<<"rest";
    if (a==4)
        cout<<"rest";
    if (a==7)
        cout<<"rest";
    
    return 0;
} 

简单的题目，疯狂判断就可以

if(n==1)
    cout<<"swim";
if(n==3)
    cout<<"program";
if(n==5)
    cout<<"read";
if(n==6)
    cout<<"math";
if(n==2||n==4||n==7)
    cout<<"rest";

当然，最后一个判断也可以写成

if(n!=1&&n!=3&&n!=5%%n!=6)
    cout<<"rest";

不过就是更麻烦些而已

#include <bits/stdc++.h> 
using namespace std;
int main()
{
    int n;
    cin >> n;
    if(n == 1)//进行4次判断，来确定输出。
    {
        cout << "swim";
    }
  
    if(n == 5) 
    {
        cout << "read";
    }
    if(n == 3) 
    {
        cout << "program";
    }
    if(n == 6)
    {
        cout << "math";
    }
    if(n == 2 or n == 4 or n == 7)//这里如果用else，会出错，因为会把rest追加到program和read末尾。
    { 
        cout << "rest";
    }
    return 0;
}

能用万能开头吗 #include <iostream>

#include <bits/stdc++.h> using namespace std; int main() { int n; cin >> n; if(n == 1)//判断各个项目，并输出指定词语 { cout << "swim"; } if(n == 3) { cout << "program"; } if(n == 5) { cout << "read"; } if(n == 6) { cout << "math"; } if(n == 2 or n == 4 or n == 7) { cout << "rest"; } return 0; }

简单判断而已

#include <bits/stdc++.h>
using namespace std;
int main()
{
int n;
cin >> n;
if (n == 1)
{
cout << "swim";
}
else if (n == 3)
{
cout << "program";
}
else if (n == 5)
{
cout << "read";
}
else if (n == 6)
{
cout << "math";
}
else
{
cout << "rest";
}
return 0;
}

#include <iostream> 
using namespace std;
int main()
{
    int n;
    cin >> n;
    if(n == 1)
    {
        cout << "swim";
    }
    if(n == 3) 
    {
        cout << "program";
    }
    if(n == 5) 
    {
        cout << "read";
    }
    if(n == 6)
    {
        cout << "math";
    }
    if(n == 2 or n == 4 or n == 7)
    { 
        cout << "rest";
    }
    return 0;
}

傻瓜式做法：

#include <iostream>
using namespace std;
int main()
{
    int n;
    cin >> n;
    if (n == 1)
    {
        cout << "swim";
    }   
    if (n == 3)
    {
        cout << "program";
    }
    if (n == 5)
    {
        cout << "read";
    }
    if (n == 6)
    {
        cout << "math";
    }
    if (n == 2 or n == 4 or n == 7)
    {
        cout << "rest";
    }
    return 0;
}

正确做法

#include <iostream>
using namespace std;
int main()
{
    cout << "I don't know!"
    return 0;
}

//这道题的话输入是一个整数,还有多种判断条件,所以建议使用switch语句
//代码: cin >> i;
switch(i){
    case 1:cout << "swim";break;//break是要有点如果没有,switch默认后面全部成立
    .../省略的代码
    ...
    ...
    default:...;break;/default表示以上全部case全部没有匹配成功的情况,一般放在最后面 } 

//这道题的话输入是一个整数,还有多种判断条件,所以建议使用switch语句 //代码: cin >> i switch(i){ case 1:cout << "swim";break;//break是要有点如果没有,switch默认后面全部成立 .../省略的代码 ... ... default:...;break;/default表示以上全部case全部没有匹配成功的情况,一般放在最后面 } </span>

这题比较简单，但有一个地方需要注意，不能用if语句判断1356四天后直接在6后加else语句，否则135三天会输出两个单词。

    if(n == 1)
    {
        cout << "swim";
    }
    if(n == 2)
    {
        cout << "rest";
    }
    if(n == 3)
    {
        cout << "program";
    }
    if(n == 4)
    {
        cout << "rest";
    }
    if(n == 5)
    {
        cout << "read";
    }
    if(n == 6)
    {
        cout << "math";
    }
    if(n == 7)
    {
        cout << "rest";
    }
    超级简单

#include <iostream> using namespace std; int main() { int a; cin >> a; if (a == 1) { cout << "swim"; } if (a == 3) { cout << "program"; } if (a == 5) { cout << "read"; } if (a == 6) { cout << "math"; } else { cout << "rest"; } return 0; }

#include <bits/stdc++.h> using namespace std; int main() { int n; cin>>n; if(n1) { cout<<"swim"; } else if(n3) { cout<<"program"; } else if(n5) { cout<<"read"; } else if(n6) { cout<<"math"; } else { cout<<"rest"; } }

题解：就是多个if条件判断，注意用一个if-else输出没有课程的内容，输出rest。 也可以使用if-else if-else。

//以下为判断部分

注：有错误发邮件告诉我 ——邮箱：liuhaoxuan_0820@hotmail.com

1. int n; cin>>n; if (n1) { cout<<"swim"; } if (n3) { cout<<"program"; } if (n5) { cout<<"read"; } if (n6) { cout<<"math"; } if (n2 or n4 or n==7) { cout<<"rest"; }

#include<bits/stdc++.h> using namespace std; int main() { int n; cin >> n; if (n == 1) { cout << "swim"; } else { if (n == 3) { cout << "program"; } else { if (n == 5) { cout << "read"; } else { if (n == 6) { cout << "math"; } else { cout << "rest"; } } } } return 0; }

这题很简单，用4次if就OK

if(n == 1)
{
cout << "swim";
}
if(n == 3)
{
cout << "program";
}
if(n == 5)
{
cout << "read";
}
if(n == 6)
{
cout << "math";
}
if(n == 2 or n == 4 or n == 7)
{
cout << "rest";
}
return 0;

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