UVA 10167 - Birthday Cake

最新推荐文章于 2018-08-12 23:17:36 发布

最新推荐文章于 2018-08-12 23:17:36 发布 · 171 阅读

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本文探讨了一个有趣的算法问题：如何通过一条直线将蛋糕上的樱桃等分为两半，确保每半都有相同数量的樱桃，同时保持蛋糕形状一致。该问题需要解决者运用随机数生成和迭代方法来找到合适的直线方程。

Background

Lucy and Lily are twins. Today is their birthday. Mother buys a birthday cake for them.Now we put the cake onto a Descartes coordinate. Its center is at (0,0), and the cake's length of radius is 100.

There are 2N (N is a integer, 1<=N<=50) cherries on the cake. Mother wants to cut the cake into two halves with a knife (of course a beeline). The twins would like to be treated fairly, that means, the shape of the two halves must be the same (that means the beeline must go through the center of the cake) , and each half must have N cherrie(s). Can you help her?

Note: the coordinate of a cherry (x , y) are two integers. You must give the line as form two integers A,B(stands for Ax+By=0), each number in the range [-500,500]. Cherries are not allowed lying on the beeline. For each dataset there is at least one solution.

Input

The input file contains several scenarios. Each of them consists of 2 parts: The first part consists of a line with a number N, the second part consists of 2N lines, each line has two number, meaning (x,y) .There is only one space between two border numbers. The input file is ended with N=0.

Output

For each scenario, print a line containing two numbers A and B. There should be a space between them. If there are many solutions, you can only print one of them.

Sample Input

Sample Output

0 1

总结问题：

1 产生a，b 用两个for，没用随机数，导致WA

2 没有用while 和n=0的判断，WA

3 数组开的太小了，WA

#define RUN
#ifdef RUN

#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <assert.h>
#include <string>
#include <iostream>
#include <sstream>
#include <map>
#include <set>
#include <vector>
#include <list>
#include <cctype> 
#include <algorithm>
#include <utility>
#include <math.h>
#include <ctime>

using namespace std;

#define MAXN 110

int input[MAXN][2];
int n;

void printout(){
	cout << "2*n:" << 2*n << endl;
	for(int i=0; i<2*n; i++){
		printf("%d %d\n", input[i][0], input[i][1]);
	}
}


void play(){
	
	int pos=0, neg=0;

	while(true){
		int a = -500 + (rand()%1001);
		int b = -500 + (rand()%1001);

		pos = 0;
		neg = 0;
		for(int k=0; k<2*n; k++){
			int sign = a*input[k][0] + b*input[k][1];
			if(sign > 0){
				pos++;
			}
			else if(sign < 0){
				neg++;
			}

			if(pos>=n && neg>=n){
				printf("%d %d\n", a, b);
				return;
			}
		}
	}
}


int main(){

#ifndef ONLINE_JUDGE
	freopen("10167.in", "r", stdin);
	freopen("10167.out", "w", stdout);
#endif

	srand ( time(NULL) );
	while(scanf("%d",&n)==1 && n!=0){
		memset(input, 0, sizeof(input));
		for(int i=0; i<2*n; i++){
			scanf("%d%d", &input[i][0], &input[i][1]);
		}
		play();
		//printout();
	}
	
	return 0;
}


#endif