Find Small A
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 3070 Accepted Submission(s): 1411
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
3 97 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:
就是转化成二进制后,每8位看是否和97的二进制一样。
有几个ans加几个。
想来自己的方法也有点愚蠢:
#include<bits/stdc++.h>
using namespace std;
const int maxn=100000;
typedef long long LL;
const int inf=0x3f3f3f3f;
int w[100];
int lala(int n){
int cnt=0;
while(n){
if(n%2==0)
w[++cnt]=0;
else
w[++cnt]=1;
n=n/2;
}
/* for(int i=1;i<=cnt;i++){
cout<<w[i]<<" ";
}
cout<<endl;
*/
return cnt;
}
int main()
{
int n,x;
scanf("%d",&n);
int num=0;
for(int i=1;i<=n;i++){
scanf("%d",&x);
int haha=lala(x);
for(int j=1;j<=haha;){///一共有haha位
int k=j;
if(k+6<=haha)
{
if( w[k]==1 &&w[k+1]==0 &&w[k+2]==0 &&w[k+3]==0 &&w[k+4]==0 &&w[k+5]==1 &&w[k+6]==1 && w[k+7]==0 )
num++;
}
j=j+8;
}
}
cout<<num<<endl;
}
简单解法:
让ans=1<<8然后用这个数取余看是否等于97,然后右移8位。
#include<bits/stdc++.h>
using namespace std;
int main()
{
int n,x;
int num;
int ans=1<<8;
while(scanf("%d",&n)!=EOF){
num=0;
for(int i=1;i<=n;i++){
scanf("%d",&x);
while(x){
if(x%ans==97)
num++;
x>>=8;
}
}
printf("%d\n",num);
}
return 0;
}