Day 10: Binary Numbers

电玩女神 2022-03-12 04:08 108阅读 0赞

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# C++: #

#include <stdio.h>
    #include <stdlib.h>
    #include <vector>
    #include <string>
    #include <algorithm>
    #include <math.h>
    #include <iostream>
    
    using namespace std;
    
    int main() {
        int n;
        cin >> n;
        cin.ignore(numeric_limits<streamsize>::max(), '\n');
    
        int a = 0; //商
        int b = 0; //余数
        int count = 0;//连续1的个数
        int Maximum = 0;//连续1的个数的最大值
        while (n != 0) {
        	a = n / 2;
    	b = n % 2;
    
    	if (b == 1) {
    		count++;
    		if (count > Maximum) {
    			Maximum = count;
    		}
    	}
    	else {
    		count = 0;
    	}
    	n = a;
        }
        cout << Maximum<<endl;
        system("pause");
        return 0;
    }

# Python: #

#!/bin/python3
    
    import math
    import os
    import random
    import re
    import sys
    
    
    
    if __name__ == '__main__':
        n = int(input())
        quot = 0 # quotient 商
        rem = 0  # remainder 余数
        count = 0 # Number of consecutive 1 连续1的个数
        Maximum = 0 # Maximum number of consecutive 's 连续1的个数的最大值
        while n != 0:
            quot = int(n / 2) # 这里要用int类型，否则默认的是float类型，出现错误
            rem = int(n % 2)  # 这里要用int类型，否则默认的是float类型，出现错误
            if rem == 1:
                count += 1
                if count > Maximum:
                    Maximum = count
            else:
                count = 0
            n = quot
        print(Maximum)

# 总结： #

## 算法思路： ##

1.  尽量循环一遍得到结果
2.  通过标记判断连续1的个数
3.  一直寻找最大值
4.  遇到相同个数的1，不必更新最大值！！！

参考：[https://blog.csdn.net/u010352111/article/details/53908250][https_blog.csdn.net_u010352111_article_details_53908250]

## C++： ##

转为二进制数，辗转相除法

while (n != 0) {
    
        a = n / 2; // 商
        b = n % 2; // 余数
    	
        n = a; // 让商作为新的除数
    }

## Python: ##

计算整除和求余数

quot = int(n / 2)  # 这里要用int类型，否则默认的是float类型，出现错误
     rem  = int(n % 2)  # 这里要用int类型，否则默认的是float类型，出现错误

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[https_blog.csdn.net_u010352111_article_details_53908250]: https://blog.csdn.net/u010352111/article/details/53908250