2016 ICPC Dalian J - Find Small A-优快云博客

本文链接：https://blog.youkuaiyun.com/chimp10/article/details/113397374

该篇博客讲述了如何通过编程解决一个计算机科学问题，即给定一组由32位整数表示的数字，计算其中与字符'a'(ASCII码为97)二进制形式相同的字符数量。代码实例展示了逐位比较的方法来找出这些字符。

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题干

 As is known to all,the ASCII of character 'a' is 97. Now,find out how many character 'a' in a group of given numbers. Please note that the numbers here are given by 32 bits’ integers in the computer.That means,1digit represents 4 characters(one character is represented by 8 bits’ binary digits).

Input
The input contains a set of test data.The first number is one positive integer N (1≤N≤100),and then N positive integersai (1≤ ai
≤2^32 - 1) follow
Output
Output one line,including an integer representing the number of ‘a’ in the group of given numbers.
Sample Input

3
97 24929 100

Sample Output

大概题意

给定一组正整数，求正整数中和’a’二进制相同的二进制码；

代码

（对位数进行操作）

#include<stdio.h>
#include<iostream>
#include<string.h>

using namespace std;
int main(){
	int n;
	cin>>n;
	int m;
	int sum=0;
	int p;
	for(int i=1;i<=n;++i){
		cin>>m;
		p=m>>24;
		if(p==97) ++sum;
		p=(m<<8)>>24;
		if(p==97) ++sum;
		p=(m<<16)>>24;
		if(p==97) ++sum;
		p=(m<<24)>>24;
		if(p==97) ++sum;
	}
	cout<<sum<<endl; 
}