nyoj 43

24 point game

题意：该题也是通过经典的24点改编的，就是给你几个数要通过 + - * / 这几个运算后（中间可以加括号），可不可以得出结果24

24点在编程之美上也有介绍，应该是出自微软面试题

该题的思路就是通过深搜枚举各种可能性：

附代码：

 
#include <stdio.h>
#include <math.h>
#include <iostream>
#include <string>
#include <algorithm>
#define Max_len 4
#define NUM 1e-6

int num;
double record[Max_len];
int getNum(int n);
using namespace std;

int main(){
	
	int N, i, k, m;
	
	scanf("%d", &N);

	while(N--){
		scanf("%d %d", &m, &num);
		for(i = 0; i < m; i++){
			scanf("%lf", &record[i]);
		}
		k = getNum(m);
		if(k)
			printf("Yes\n");
		else{
			printf("No\n");
		}
	}

	return 0;
}	

int	getNum(int n){
	
	int i, j;
	double a, b;
	
//	printf("%d\n", n);
	if(n == 1){
		if(fabs(record[0] - num) < NUM){
			return 1;
		}
		//printf("%lf", record[0]);
	//	getchar();
		return 0;
	}

	for(i = 0; i < n; i++){
		for(j = i + 1; j < n; j++){
			a = record[i]; b = record[j];
			record[j] = record[n - 1];

			record[i] = a + b;
			if(getNum(n - 1))
				return 1;
			

			record[i] = a - b;
			if(getNum(n - 1))
				return 1;

			record[i] = a * b;
			if(getNum(n - 1))
				return 1;

			record[i] = b - a;
			if(getNum(n - 1))
				return 1;
			
			if(b != 0){
				record[i] = a / b;
				if(getNum(n - 1))
					return 1;
			}
		
			if(a != 0){
				record[i] = b / a;
				if(getNum(n - 1))
					return 1;
			}

			record[i] = a;
			record[j] = b;
		}
	}

	return 0;
}        

nyoj 43

原文：http://blog.csdn.net/huntinggo/article/details/19424407