题目描述

这道题有个连除规律，用循环解确实更简洁些，就是输出格式有点复杂，粘贴的时候容易出错

样例

#include<cstdio>
using namespace std;
int main(){
    int N;
    scanf("%d",&N);
    printf("%d\n",N);
    printf("%d nota(s) de R$ 100,00\n",N/100);
    N=N%100;
    printf("%d nota(s) de R$ 50,00\n",N/50);
    N=N%50;
    printf("%d nota(s) de R$ 20,00\n",N/20);
    N=N%20;
    printf("%d nota(s) de R$ 10,00\n",N/10);
    N=N%10;
    printf("%d nota(s) de R$ 5,00\n",N/5);
    N=N%5;
    printf("%d nota(s) de R$ 2,00\n",N/2);
    N=N%2;
    printf("%d nota(s) de R$ 1,00\n",N);
    return 0;
}

算法1

(暴力枚举) $O(n^2)$

时间复杂度

参考文献

C++ 代码

#include<iostream>
using namespace std;
int main(){
    int N;
    cin>>N;
    cout<<N<<endl;
    cout<<N/100<<" "<<nota(s) de R$ 100,00\n<<endl);
    N=N%100;
    cout<<N/50<<" "<<nota(s) de R$ 50,00\n<<endl;
    N=N%50;
    cout<<N/20<<" "<<nota(s) de R$ 20,00\n<<endl;;
    N=N%20;
    cout<<N/10<<" "<<nota(s) de R$ 10,00\n<<endl;
    N=N%10;
    cout<<N/5<<" "<<nota(s) de R$ 5,00\n<<endl;
    N=N%5;
    cout<<N/2<<" "<<nota(s) de R$ 2,00\n<<endl;
    N=N%2;
    cout<<N/1<<" "<<nota(s) de R$ 1,00\n<<endl;
    return 0;
}

算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla