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).
InputThe 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) followOutputOutput 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
1 #include <iostream> 2 #include <cstdio> 3 using namespace std; 4 5 int main() 6 { 7 int ans,n,x; 8 int i; 9 cin>>n; 10 ans=0; 11 for(i=0;i<n;i++) 12 { 13 scanf("%d",&x); 14 while(x) 15 { 16 if(x%256==97) 17 { 18 ans++; 19 } 20 x/=256; 21 } 22 } 23 cout<<ans<<endl; 24 }
时间: 2024-10-02 18:51:23