CodeForces 742B Arpa’s obvious problem and Mehrdad’s terrible solution (暴力枚举)
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题意:求定 n 个数,求有多少对数满足,ai^bi = x。
析:暴力枚举就行,n的复杂度。
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> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e5 + 5; const int mod = 1e9 + 7; const int dr[] = {-1, 1, 0, 0}; const int dc[] = {0, 0, 1, -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 int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int a[100005]; int b[100005]; int main(){ LL n, m; while(scanf("%I64d %I64d", &n, &m) == 2){ for(int i = 0; i < n; ++i) scanf("%d", a+i); memset(b, 0, sizeof b); LL ans = 0; // if(m == 0){ // for() // continue; // } for(int i = n-1; i >= 0; --i){ LL t = a[i] ^ m; if(t < 100005) ans += b[t]; ++b[a[i]]; } cout << ans << endl; } return 0; }
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