首页 > 代码库 > HDU 5980 Find Small A (水题)
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)
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。