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  • 2016 ICPC Dalian-Find Small A

    Find Small A(点我可传送)

    题面

    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 integers (ai(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

    3

    题目分析

    题意:

    众所周知,在ASCII码中,a的值为97。现在给你一组整数,请你计算里面有多少个a

    请注意,所给的数字在计算机中都是32位整数,而一个字符占8位


    解题思路:

    本题只需要判断每一个数字(x),然后“8位8位”的看——每次只看二进制中的后八位(也就是 x % (2^8) ),检查这8位中是否有一个a,即x % 256 == 97。当前后8位判完后,再右移8位(即除以(2^8)),判断下一组8位,直到x == 0

    AC代码

    #include <bits/stdc++.h>
    using namespace std;
    int n;
    int main() {
        ios::sync_with_stdio(false), cin.tie(0), cout.tie(0);
        __int32 x;
        while (cin >> n) {
            int ans = 0;
            for (int i = 0; i < n; i++) {
                cin >> x;
                while (x) {
                    if (x % 256 == 'a')
                        ans++;
                    x >>= 8;
                }
            }
            cout << ans << endl;
        }
    
        return 0;
    }
    
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  • 原文地址:https://www.cnblogs.com/FrankOu/p/14346353.html
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