HDU 5980 - Find Small A（进制）_problem description as is known to all,the ascii o-优快云博客

本文介绍了一个编程问题，任务是计算一组32位整数中字符'a'出现的次数。通过将整数转换为二进制并检查每8位是否等于字符'a'的ASCII值97来实现。

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Find Small A

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1680 Accepted Submission(s): 826

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）

给你n个十进制的数，让你把它变成2进制，然后每8位算一下是不是等于97，是就ans++。

#include <iostream>
#include <string.h>
#include <stdio.h>
#include <algorithm>
#include <queue>
#include <math.h>
#include <vector>
using namespace std;
const int maxn = 55;
int a[maxn];
int ans;
int num=0;
void er(int x)
{
    num=0;
    while(x){
        num++;
        a[num]=x%2;
        x=x/2;
    }
}
int main()
{
    int n;
    while(~scanf("%d",&n)){
        ans=0;
        for(int i=1;i<=n;i++){
            int p;scanf("%d",&p);
            er(p);
            int cnt=0;
            int now=0;
            for(int j=1;j<=num;j++){
                cnt++;
                if(a[j]){
                    now+=1<<(cnt-1);
                }
                if(cnt==8){
                    if(now==97){
                        ans++;
                    }
                    now=0;
                    cnt=0;
                }
            }
            if(now==97){
                ans++;
            }
        }
        printf("%d\n",ans);
    }
}