题目描述

这道题还是有点难度的需要掌握贪心算法

样例

#include<iostream>
using namespace std;
int main() {
    int N;
    cin >> N;
    cout<<N<<endl;
    const int qianleizongshu = 7;
    int qianzhong[qianleizongshu] = { 100,50,20,10,5,2,1 };
    int meizhongqianshuliang[qianleizongshu] = { 0 };
    for (int i = 0; i < qianleizongshu && N>0; i++) {
        meizhongqianshuliang[i] = N / qianzhong[i];
        N -= meizhongqianshuliang[i] * qianzhong[i];
    }

    for (int i = 0; i < qianleizongshu; i++) {

            cout << meizhongqianshuliang[i] << " nota(s) de R$ " << qianzhong[i] <<",00"<< endl;


    }
    return 0;
}

算法1

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

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算法2

(暴力枚举) $O(n^2)$

blablabla

时间复杂度

参考文献

C++ 代码

blablabla