米老鼠怎么开始偷糖果了呢

很简单，cin+cout：

#include <bits/stdc++.h>
using namespace std;

int main()
{
//  freopen(" .in", "r", stdin);
//  freopen(" .out", "w", stdout);
    int n, a, x;
    cin >> n >> a >> x;
    cout << n-a*x;
    return 0;
}

米老鼠偷糖果

#include <iostream> using namespace std; int main() { int a,x,n; cin >> n >> a >> x; cout << n-a*x; return 0; }

题解：简单数学知识，n-a*x，计算剩余数量。

这题太简单了所已直接上代码 #include<bits/stdc++.h> using namespace std; int main(){ int n=0,a=0,x=0; cin>>n>>a>>x; cout<<n-a*x; return 0; }

#include <iostream>
using namespace std;
int main()
{
    int n,a,x;
    cin>>n>>a>>x;
    cout<<n-a*x;
    return 0;
}
* ```none
  //彩蛋
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  {      /~~\  :--~""""\.:  :
   \    (... :   /^\  /^\\ ;
    ~\_____     |   | | |:~
          /     |__O|_|O|;
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           :               ||
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Copy *

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int n, a, x;
    //分别是糖果总数、每次运的颗数和运的次数三   个变量
    cin >> n >> a >> x;
    //输入三个变量
    n = n - a * x;
    //从总数中减去运的总颗数
    cout << n;
    //输出最后结果
    return 0;
}

#include <iostream> using namespace std; int main() { int n,a,x; cin>>n>>a>>x; cout<<n-(a*x); return 0; }

这道题真的好难 讲真的，这题没什么好说的 对了，下面的代码有一点小错误，请自己找出并改正😄

#include <bit/stdc++.h>
using namespace std;
int a,b,c;
int main()
{
    cin >> a>>b>>c;
    cout <<a-b*c;
}

#include <iostream>
using namespace std;
int main()
{
    **存仨变量;**
    把一个当输入用;
    存储十位;
    存储个位;
    如果甲比乙大
    {
        输出甲;
    }

    否则
    {
        输出乙;
    }

    return 0;
}//点赞，点赞！！！

废话不多说，上代码。

#include<iostream>
using namespace std;
int main()
{
    int n,a,x;
    cin>>n>>a>>x;
    for(int i=1;i<=x;i++)
    {
        n-=a;
    }
    cout<<n;
    return 0;
}

不会吧，这么简单 应该都会吧......(先赞 后看！养成习惯！)

#include<iostream>
using namespace std;
int n,a,x;
int main()
{
    cout<<5;
    //相信我，绝对行
    //绝对AC
    return 0;
}

制作不易,给个赞吧,球球了...... 有什么问题,联系我,邮箱是ASheepBoy_Bed@163.com

#include <iostream>
using namespace std;
int main()
{
int n,a,x;//定义
cin >>n>>a>>x;//输入
cout <<n-x*a;//输出用总糖果数减去偷的糖果数（奇奇怪怪的米老鼠）
return 0;
}

来力！！！！ python3

p=int(input())
o=int(input())
i=int(input())
o=o*i
p=p-o
print(p)

jjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjjj

非常简单

#include <bits/stdc++.h>
using namespace std;
int main()
{
    int a,n,x,sum=0;
    cin >> n >> a >> x  ;
    sum=n-a*x;
    cout<<sum;
    return 0;
}

#include <bits/stdc++.h> using namespace std; int main() { int n, a, x; cin >> n >> a >> x; cout << n - a * x;

return 0;

}

有手就行

#include <bits/stdc++.h> using namespace std; int main() { int n,a,x; cin >> n >> a >> x; cout << n - a * x;//总量-每次数*次数 return 0; }