题意：

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≤ aiai≤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

3

思路：

​ int类型的数在二进制下有32位，问你一个数组中的数，8位8位的看，问存在多少个97。因此我们每个数累除256直到0，每次累除都取余256看是否存在97，存在就++即可

代码：

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

int main () {
    int n;
    while(scanf("%d", &n) == 1) {
        unsigned int a;
        int sum =0;
        for (int i = 1; i <= n; i++){
            scanf("%u", &a);
            while(a) {
                if (a % 256 == 97) sum ++;
                a /= 256;
            }
        }
        printf("%d\n", sum);
    }
    return 0;
}

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