浙江财经大学
信息管理与工程学院

HDU 2700 Parity(瞎搞题)

本文由 Ocrosoft 于 2016-05-23 21:49:29 发表

Parity

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 4200    Accepted Submission(s): 3158

Problem Description
A bit string has odd parity if the number of 1’s is odd. A bit string has even parity if the number of 1’s is even.Zero is considered to be an even number, so a bit string with no 1’s has even parity. Note that the number of
0’s does not affect the parity of a bit string.
 

Input
The input consists of one or more strings, each on a line by itself, followed by a line containing only “#” that signals the end of the input. Each string contains 1–31 bits followed by either a lowercase letter ‘e’ or a lowercase letter ‘o’.
 

Output
Each line of output must look just like the corresponding line of input, except that the letter at the end is replaced by the correct bit so that the entire bit string has even parity (if the letter was ‘e’) or odd parity (if the letter was ‘o’).
 

Sample Input
	
101e 010010o 1e 000e 110100101o #
 

Sample Output
	
1010 0100101 11 0000 1101001010
 

Source
 

Solution

题意:一个二进制串中有奇数个1则认为它是奇数,偶数个1则为偶数。

如果二进制串结尾是o,则使其成为奇数。如果结尾是e,则使其成为偶数。

题解:统计1的个数,结尾是e时,如果1是偶数个的,e变成0,否则变成1.

结尾是o时,如果1是奇数个,o变成0,否则变成1.

 
#include <iostream>
using namespace std;
int main()
{
    string s;
    while(cin>>s&&s.compare("#"))
    {
        int cnt=0;
        for(string::iterator it=s.begin();it!=s.end();it++)if(*it=='1')cnt++;
        if(cnt%2)*(s.end()-1)=*(s.end()-1)=='e'?'1':'0';
        else *(s.end()-1)=*(s.end()-1)=='e'?'0':'1';
        cout<<s<<endl;
    }
    return 0;
}

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