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HDU 5980 Find Small A (水题)

题意:众所周知,字符 ‘a‘ 的ASCII码为97.现在,找出给定数组中出现了多少次 ‘a‘ 。注意,此处的数字为计算机中的32位整数。这表示,

1个数字由四个字符组成(一个字符由8位二进制数组成)。

析:直接用运用位运算即可。

代码如下:

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <stack>
#include <sstream>
#define debug() puts("++++");
#define gcd(a, b) __gcd(a, b)
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;

typedef long long LL;
typedef unsigned long long ULL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const LL LNF = 1e16;
const double inf = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 1e5 + 10;
const int mod = 1e9 + 7;
const int dr[] = {-1, 0, 1, 0};
const int dc[] = {0, 1, 0, -1};
const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline bool is_in(int r, int c){
  return r >= 0 && r < n && c >= 0 && c < m;
}

int main(){
  string str;
  for(int i = 0; i < 8; ++i)
    str.push_back((97&(1<<i)) ? ‘1‘ : ‘0‘);
  int T;  cin >> T;
  int ans = 0;
  while(T--){
    string s;
    LL n;
    cin >> n;
    for(int i = 0; i < 32; ++i)
      s.push_back((n&(1LL<<i)) ? ‘1‘ : ‘0‘);
    for(int i = 0; i < 32; i += 8)
      if(str == s.substr(i, 8))  ++ans;
  }
  cout << ans << endl;
  return 0;
}

  

HDU 5980 Find Small A (水题)