NYOJ-43 Point Game-dfs

Point game time limit: theMs | Memory Limit:65535KB Difficulty:5

Describe

There is a game which are called point game.

In this game, you'll be given some numbers. Your task is to find a expression which the given numbers and the value of the expression should be 24. The expression mustn ' t has any other operator except Plus,minus,multiply,divide and the brackets.

e.g. If the numbers given is "3 3 8 8", you can Give "8/(3-8/3)" as an answer. All the numbers should is used and the bracktes can be nested.

Your task in this problem are only to judge whether the given numbers can being used to find a expression whose value is the G Iven number.

Input

The input has multicases and each case contains one line
The first line of the input was an non-negative integer C (c<=100), which indicates the number of the cases.
Each line have some integers,the first integer M (0<=m<=5) is the total number of the given numbers to consist the EXP ression,the second integers N (0<=n<=100) is the number which the value of the expression should be.
Then,the followed M integer is the given numbers. All the given numbers are non-negative and less than 100

Output

For each

test-cases,output ' Yes ' if there is a expression which fit all the demands,otherwise output "No" instead.

Sample input

24 24 3 3 8 83 24 8 3 3

Sample output

YesNo

#include <cstdio> #include <cstring> #include <cmath> #include <cstdlib> #include <iostream > #include <algorithm> #include <queue> #include <stack>//#define MIN 1e-10using namespace std; Const double Min=1e-10;//const double, habitually written as const int leads to a night double a[10]; int n;double num;bool DFS (int x) {if (x==n) {if (Fabs (Num-a[n]) <min) return True;return false; Double now,next;for (int i=x;i<n;i++) {for (int j=i+1;j<=n;j++) {now=a[i];next=a[j];a[i]=a[x];a[j]=now+next;if ( DFS (x+1)) return True;a[j]=now-next;if (Dfs (x+1)) return True;a[j]=next-now;if (Dfs (x+1)) return true;a[j]=now*next;if (Dfs (x+1)) return true;if (Next)//denominator not 0 {a[j]=now/next;if (DFS (x+1)) return true;} if (now)//The denominator is not 0{a[j]=next/now;if (DFS (x+1)) return true; A[i]=now;a[j]=next;}} return false;} int main () {int t;while (~scanf ("%d", &t)) {while (t--) {memset (a,0,sizeof (a)); scanf ("%d%lf", &n,&num); for ( int i=1;i<=n;i++) {scanf ("%lf", &a[i]);} if (DFS (1)) cout<< "Yes" <<endl; else cout<< "No" <<endl;}} return 0;}

NYOJ-43 Point Game-dfs