CodeForces 514B – Watto and Mechanism

这题看网上题解,都是相除然后什么暴力的,感觉没有必要。实际上就是计算每个点相当于炮塔的斜率,考虑到斜率是浮点数,因此可以把分数约分,然后比较有多少个这样不同的最简分数即可。刚开始用数组开的比较大内存超限了,后来改用map辅助pair占用空间就少了。 

Description

Watto, the owner of a spare parts store, has recently got an order for the mechanism that can process strings in a certain way. Initially the memory of the mechanism is filled with n strings. Then the mechanism should be able to process queries of the following type: "Given strings, determine if the memory of the mechanism contains string t that consists of the same number of characters as s and differs from s in exactly one position".

Watto has already compiled the mechanism, all that's left is to write a program for it and check it on the data consisting of n initial lines andm queries. He decided to entrust this job to you.

 

Input

The first line contains two non-negative numbers n and m (0 ≤ n ≤ 3·1050 ≤ m ≤ 3·105) — the number of the initial strings and the number of queries, respectively.

Next follow n non-empty strings that are uploaded to the memory of the mechanism.

Next follow m non-empty strings that are the queries to the mechanism.

The total length of lines in the input doesn't exceed 6·105. Each line consists only of letters 'a''b''c'.

 

Output

For each query print on a single line "YES" (without the quotes), if the memory of the mechanism contains the required string, otherwise print "NO" (without the quotes).

 

Sample Input

Input
2 3
aaaaa
acacaca
aabaa
ccacacc
caaac

 

Output
YES
NO
NO

 

 

Source

#include <bits/stdc++.h>
using namespace std;


set<pair<int, int>> s;
int gcd(int a, int b) {
    return b ? gcd(b, a % b) : a;
}

int main() {
    int n, x, y; cin >> n >> x >> y;
    int cnt = 0;
    while(n--){
        int r, c; cin >> r >> c;
        r -= x; c -= y;
        int g = gcd(r, c);
        r /= g; c /= g;
        if(!s.count(make_pair(r, c))){
            cnt++;
            s.insert(make_pair(r, c));
        }
    }
    cout << cnt << endl;
    return 0;
}

 

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