hdu-5980-Find Small A

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最新推荐文章于 2022-01-02 19:02:52 发布

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本文介绍了一种计算一组32位整数中字符'a'出现次数的方法。通过将整数转换为二进制并分段检查每段是否等于字符'a'的ASCII值97对应的二进制形式。

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题目链接：http://acm.split.hdu.edu.cn/showproblem.php?pid=5980

Find Small A

Problem Description

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

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）

将一个32位int型的整数换成32位的2进制，可以拆成4部分，问这四部分里面出现了多少次97的二进制1100001，可以将目标数每次对256取余，取余结果是多少就表示出现了多少次97，将该数左移8位继续取余操作。

#include<iostream>
#include<cmath>
#include<iomanip>
#include<cstdio>
using namespace std;
typedef long long ll;
const double pi=acos(-1.0);
int main(){
	int n;
	ll x;
	while(cin>>n){
		int ans=0;
		for(int i=0;i<n;i++){
			cin>>x;
			for(int j=0;j<4;j++){
				if(x%256==97)
				ans++;
				x>>=8;
			}
		}
		cout<<ans<<endl;
	}	
	return 0;
}