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  • 【cf492】E. Vanya and Field(拓展欧几里得)

    http://codeforces.com/contest/492/problem/E

    一开始没时间想,,诶真是。。

    挺水的一道题。。

    将每个点的横坐标都转换成0,然后找纵坐标有多少即可。。即解方程

    $$a imes dx equiv x(mod n)$$

    然后注意开long long

    #include <cstdio>
    #include <cstring>
    #include <cmath>
    #include <string>
    #include <iostream>
    #include <algorithm>
    #include <queue>
    #include <set>
    #include <map>
    using namespace std;
    typedef long long ll;
    #define rep(i, n) for(int i=0; i<(n); ++i)
    #define for1(i,a,n) for(int i=(a);i<=(n);++i)
    #define for2(i,a,n) for(int i=(a);i<(n);++i)
    #define for3(i,a,n) for(int i=(a);i>=(n);--i)
    #define for4(i,a,n) for(int i=(a);i>(n);--i)
    #define CC(i,a) memset(i,a,sizeof(i))
    #define read(a) a=getint()
    #define print(a) printf("%d", a)
    #define dbg(x) cout << (#x) << " = " << (x) << endl
    #define error(x) (!(x)?puts("error"):0)
    #define rdm(x, i) for(int i=ihead[x]; i; i=e[i].next)
    inline const int getint() { int r=0, k=1; char c=getchar(); for(; c<'0'||c>'9'; c=getchar()) if(c=='-') k=-1; for(; c>='0'&&c<='9'; c=getchar()) r=r*10+c-'0'; return k*r; }
    
    const int N=1e6+10;
    void exgcd(ll a, ll b, ll &d, ll &x, ll &y) {
    	if(!b) { d=a; x=1; y=0; return; }
    	exgcd(b, a%b, d, y, x); y-=a/b*x;
    }
    int n, m, c[N];
    ll x, y, d, a, b, dx, dy;
    int main() {
    	cin >> n >> m >> dx >> dy;
    	for1(i, 1, m) {
    		cin >> x >> y;
    		if(x==0) { c[y]++; continue; }
    		exgcd(dx, n, d, a, b);
    		if(x%d!=0) continue;
    		a=(a+n)%n;
    		a=(a*x)%n;
    		x=((x-a*dx)%n+n)%n;
    		y=((y-a*dy)%n+n)%n;
    		c[y]++;
    	}
    	int mx=0, pos=-1;
    	rep(i, n) if(c[i]>mx) pos=i, mx=c[i];
    	//rep(i, n) dbg(c[i]);
    	if(pos==-1) cout << x << ' ' << y << endl;
    	else cout << 0 << ' ' << pos << endl;
    	return 0;
    }
    

      


    Vanya decided to walk in the field of size n × n cells. The field contains m apple trees, the i-th apple tree is at the cell with coordinates (xi, yi). Vanya moves towards vector (dx, dy). That means that if Vanya is now at the cell (x, y), then in a second he will be at cell . The following condition is satisfied for the vector: , where  is the largest integer that divides both a and b. Vanya ends his path when he reaches the square he has already visited.

    Vanya wonders, from what square of the field he should start his path to see as many apple trees as possible.

    Input

    The first line contains integers n, m, dx, dy(1 ≤ n ≤ 106, 1 ≤ m ≤ 105, 1 ≤ dx, dy ≤ n) — the size of the field, the number of apple trees and the vector of Vanya's movement. Next m lines contain integers xi, yi (0 ≤ xi, yi ≤ n - 1) — the coordinates of apples. One cell may contain multiple apple trees.

    Output

    Print two space-separated numbers — the coordinates of the cell from which you should start your path. If there are several answers you are allowed to print any of them.

    Sample test(s)
    input
    5 5 2 3
    0 0
    1 2
    1 3
    2 4
    3 1
    output
    1 3
    input
    2 3 1 1
    0 0
    0 1
    1 1
    output
    0 0
    Note

    In the first sample Vanya's path will look like: (1, 3) - (3, 1) - (0, 4) - (2, 2) - (4, 0) - (1, 3)

    In the second sample: (0, 0) - (1, 1) - (0, 0)

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  • 原文地址:https://www.cnblogs.com/iwtwiioi/p/4136446.html
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