HDU 5980 Find Small A（对二进制理解）

Find Small A

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 427    Accepted Submission(s): 220

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

 

 题意：给N个数 每个数都可以拆开成一个32位的2进制 每八位一个字节  每个字节的2进制数换算成十进制的看有多少个97

思路：CillyB： % 256是看看后8位是多少， /= 256是去掉后8位，仔细想想 % 2^8其实就是后面8位的值（这一点记住吧） /= 就是 >> 8位

#include<iostream>#include<cstdio>#include<cstring>using namespace std;int main(void){    int n;    while(cin >> n)    {        int ans = 0, t;        for(int i = 0; i < n; i++)        {            scanf("%d", &t);            while(t)            {                if(t%256 == 97) ans++;                t /= 256;            }        }        printf("%d\n", ans);    }    return 0;}