HDU 5980 Find Small A（水题）

Find Small A

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)

Total Submission(s): 86    Accepted Submission(s): 50

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

397 24929 100

 

Sample Output

3

 

Source

2016ACM/ICPC亚洲区大连站-重现赛（感谢大连海事大学）  

题意：给出一个数字代表一个4位的字符（讲真一开始我怎么读都没明白），就是说一个数可以用32位二进制表示，每8位代表一个字符（ASCII码），问给出的数字里面包含多少个A（即ASCII码为97）。

思路：明白题意之后一开始还是想复杂了，想把一个数字转化为二进制再做，但后来发现只要直接移位然后清高位保留低八位判断是否等于97即可。WA了一次是因为中间循环条件的设置，不应该判断是否为0，因为有可能正好有一个数字中间有某个8位恰好为0。因此还是老老实实移32位之后再跳出保险。

#include <iostream>#include <string.h>#include <algorithm>#include <stdio.h>#include <math.h>using namespace std;int main(){int n;while(~scanf("%d",&n)){long long cnt=0;long long p=pow(2,8)-1;for(int i=0;i<n;i++){long long d;scanf("%lld",&d);long long temp=d;int j=0;while(j<=4){int k=8*j;j++;temp=d>>k;temp=temp&p;if(temp==97)cnt++;}}printf("%lld\n",cnt);}return 0;}