AtCoder Grand Contest 025 Problem D

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AtCoder Grand Contest 025 Problem D

　　Time Limit: 2 Sec
　　Memory Limit: 1024 MB

Description

　　Takahashi is doing a research on sets of points in a plane. Takahashi thinks a set \(S\) of points in a coordinate plane is a good set when \(S\) satisfies both of the following conditions:
　　? The distance between any two points in \(S\) is not \(\sqrt{D1}\).
　　? The distance between any two points in \(S\) is not \(\sqrt{D2}\)?.
　　Here, \(D1\) and \(D2\) are positive integer constants that Takahashi specified.
　　Let \(X\) be a set of points \((i,j)\) on a coordinate plane where \(i\) and \(j\) are integers and satisfy \(0≤i,j<2N\).
　　Takahashi has proved that, for any choice of \(D1\) and \(D2\), there exists a way to choose \(N^2\) points from \(X\) so that the chosen points form a good set. However, he does not know the specific way to choose such points to form a good set. Find a subset of \(X\) whose size is \(N^2\) that forms a good set.
　

Input

　　? \(1≤N≤300\)
　　? \(1≤D1≤2×105\)
　　? \(1≤D2≤2×105\)
　　? All values in the input are integers.
　
　　Input is given from Standard Input in the following format:
　　\(N\) \(D_1\) \(D_2\)
　　

Output

　　Print N2 distinct points that satisfy the condition in the following format:
　　\(x_1 y_1\)
　　\(x_2 y_2\)
　　:
　　\(x_{N^2} y_{N^2}\)
　　Here, (xi,yi) represents the i-th chosen point. \(0≤xi,yi<2N\) must hold, and they must be integers. The chosen points may be printed in any order. In case there are multiple possible solutions, you can output any.
　　

Sample Input 1

　　2 1 2
　

Sample Output 1

　　0 0
　　0 2
　　2 0
　　2 2
　　

Sample Input 2

　　3 1 5
　

Sample Output 2

　　0 0
　　0 2
　　0 4
　　1 1
　　1 3
　　1 5
　　2 0
　　2 2
　　2 4
　　

题目地址：　　AtCoder Grand Contest 025 Problem D

题目大意：

　　输? \(n, d_1, d_2\)，
　　你要找到 \(n^2\) 个整点 \(x, y\) 满? \(0 ≤ x, y < 2n\)。 并且找到的任意两个点距离，既不是 \(\sqrt{d1}，也不是 \sqrt{d2}\)。
　　

题解：

　　这是个分析题?。
　　简单来说，所有距离为 \(\sqrt{d_1}\) 的点连边，可以得到?个?分图。\(d_2\) 同理。
　　这样可以把所有 \(4n^2\) 个点四分，?定有?块满?条件。
　　如果 d mod 2 = 1，如果 \(a^2 + b^2 = d\)，a 和 b ?定?奇?偶，按国际象棋??染?即可。
　　如果 d mod 4 = 2，如果 \(a^2 + b^2 = d\)，a 和 b ?定都是奇数，????，????即可。
　　如果 d mod 4 = 0，把 2 × 2 的区域看成?个?格?，如此类推，对 d/4 进?如上考虑即可。
　　

#include <cstdio>
using namespace std;
int n,d1,d2,s;
int f[620][620];
void work(int d){
    int p=0;
    while(d%4==0){
        d/=4;
        p++;
    }
    if(d&1){
        for(int i=0;i<2*n;i++)
            for (int j=0;j<2*n;j++)
                if(((i>>p)+(j>>p))&1)
                    f[i][j]=1;
    }else{
        for(int i=0;i<2*n;i++)
            for(int j=0;j<2*n;j++)
                if((i>>p)&1)
                    f[i][j]=1;
    }
}
int main(){
    scanf("%d%d%d",&n,&d1,&d2);
    work(d1);
    work(d2);
    for(int i=0;i<2*n;i++){
        for(int j=0;j<2*n;j++){
            if(s<n*n && !f[i][j]){
                printf("%d %d\n",i,j);
                s++;
            }
        }
    }
    return 0;
}

原文地址：https://www.cnblogs.com/shaokele/p/9262229.html