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CodeForces 742B Arpa’s obvious problem and Mehrdad’s terrible solution (暴力枚举)
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;
}
CodeForces 742B Arpa’s obvious problem and Mehrdad’s terrible solution (暴力枚举)
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