uva10344 23 out of 5

Your task is to write a program that can decide whether you can nd an arithmetic expression consisting
of ve given numbers ai (1 i 5) that will yield the value 23.
For this problem we will only consider arithmetic expressions of the following from:
(((a(1) o1 a(2)) o2 a(3)) o3 a(4)) o4 a(5)
where : f1;2;3;4;5g ! f1;2;3;4;5g is a bijective function and oi 2 f+; ;g(1 i 4)
Input
The Input consists of 5-Tupels of positive Integers, each between 1 and 50.
Input is terminated by a line containing ve zero's. This line should not be processed. Input le
will have no more than 25 lines.
Output
For each 5-Tupel print `Possible' (without quotes) if their exists an arithmetic expression (as described
above) that yields 23. Otherwise print `Impossible'.
Sample Input
1 1 1 1 1
1 2 3 4 5
2 3 5 7 11
0 0 0 0 0
Sample Output
Impossible
Possible
Possible

题目大意如下：给定5个正整数，和一个算术集合{+，-，*}，求23点。

思路：因为24点或23点或n点的计算公式为(((a(1) o1 a(2)) o2 a(3)) o3 a(4)) o4 a(5)，可知需要在5个正整数中选定一个初始的数来作为基础值，所以枚举此数，再进行回溯。

6748808

ksq2013

UVA

10344

Accepted

470

C++11 5.3.0

698

2016-08-04 20:58:14

#include<cstdio>
#include<cstring>
#include<iostream>
using namespace std;
int a[6];
bool vis[6];
bool read()
{
	bool useable=0;
	for(int i=0;i<5;i++){
		scanf("%d",&a[i]);
		if(a[i])useable=1;
	}
	return useable;
}
bool dfs(int cur,int ans)
{
	if(cur==4&&ans==23)return 1;
	for(int i=0;i<5;i++)
		if(!vis[i]){
			vis[i]=1;
			if(dfs(cur+1,ans+a[i]))return 1;
			if(dfs(cur+1,ans*a[i]))return 1;
			if(dfs(cur+1,ans-a[i]))return 1;
			vis[i]=0;
		}
	return 0;
}
bool solve()
{
	for(int i=0;i<5;i++){
		memset(vis,0,sizeof(vis));
		vis[i]=1;
		if(dfs(0,a[i]))return 1;
		vis[i]=0;
	}
	return 0;
}
int main()
{
	while(read()){
		if(solve())puts("Possible");
		else puts("Impossible");
	}
	return 0;
}