10344-23 out of 5, intelligence games:

10344-23 Out of 5

7004

31.42%

1882

73.33%

Question link:

Http://uva.onlinejudge.org/index.php? Option = com_onlinejudge & Itemid = 8 & category = 108 & page = show_problem & problem = 1285

Type: backtracking

Original question:

Your task is to write a program that can decide whether you can find an arithmetic expression consisting of five given numbers (1 <= I <= 5)
That will yield the value 23.
For this problem we will only consider arithmetic expressions of the following from:

where : {1,2,3,4,5} -> {1,2,3,4,5} is a bijective function

and  {+,-,*} (1<=i<=4)

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

Question:

The meaning of this question is a good question. The game rules are basically the same as the "24 o'clock" of playing cards.

The rule is as follows: Give you five cards and use the +,-, and * symbols to calculate the five numbers so that they are 23.

You need to program to determine whether all five cards can be implemented.

Ideas and summary:

This game is believed to have been played by many people when they were young. If we could compile a small program to help us when we were young, then ...... It seems that programming is still a bit useful,

At least you can lie to a child...

If we use a computer and Recursive Backtracking Algorithm to do this, it will be very simple. We only need to find all the arrays and use all the symbols,

You can easily get the answer

Code implementation:

# Include <iostream> # include <cstdio> # include <cstring> using namespace STD; int arr [5]; bool vis [5], flag; void DFS (INT cur, int sum) {If (FLAG) return; // once the solution is found, you can continue to roll back the stack for (INT I = 0; I <5; ++ I) if (! Vis [I]) {vis [I] = true; If (! Cur) DFS (cur + 1, arr [I]); else {DFS (cur + 1, Sum-Arr [I]); // subtraction DFS (cur + 1, sum + arr [I]); // Add DFS (cur + 1, sum * arr [I]); // multiplication} vis [I] = false; // backtracking} If (cur = 5 & sum = 23) {flag = true; return ;}} int main () {# ifdef local freopen ("input.txt ", "r", stdin); # endif while (~ Scanf ("% d", & arr [0], & arr [1], & arr [2], & arr [3], & amp; arr [4]) {If (! Arr [0] &! Arr [1] &! Arr [2] &! Arr [3] &! Arr [4]) break; flag = false; memset (VIS, 0, sizeof (VIS); DFS (0, 0); If (FLAG) printf ("Possible \ n"); else printf ("impossible \ n");} return 0 ;}
-- The meaning of life is to give it meaning.
 
  
   
   Original
   Http://blog.csdn.net/shuangde800
   , By d_double(For details, refer)