Uva10344-23 out of 5

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)

Input

The input consists of 5-tupels of positive integers, each between 1 and 50.
Input is terminated by a line containing five zero's. This line shoshould not be processed.

Output

For each 5-tupel print "possible" (without quotes) if their exists an arithmetic expression (as described abve) 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

 

// Question: enter 5 integers and sort them in a certain order and then perform the plus (+), minus (-), or minus (*) to make the final result 23. Determine whether there is a solution
// Algorithm: backtracking

Time 1.692

# Include <cstdio> # include <cstring> # include <iostream> # include <string> # include <algorithm> using namespace STD; int A [5]; int B [5]; char op [5]; int vis [5]; int possible; void DFS (INT D, int s) {If (D = 5) {If (S = 23) {Possible = 1; # ifndef online_judge printf ("(% d", B [0]); For (INT I = 1; I <4; I ++) printf ("% C % d)", op [I], B [I]); printf ("% C % d ", OP [4], B [4]); printf ("\ n"); # endif} return;} For (INT I = 0; I <5; I ++) {// The first number if (D = 0) {If (! Vis [I]) {vis [I] = 1; B [d] = A [I]; DFS (D + 1, a [I]); vis [I] = 0 ;}} else {If (! Vis [I]) {vis [I] = 1; B [d] = A [I]; OP [d] = '+'; DFS (D + 1, S + A [I]); OP [d] = '-'; DFS (D + 1, S-A [I]); OP [d] = '*'; DFS (D + 1, S * A [I]); vis [I] = 0 ;}}} int main () {# ifndef online_judge freopen (". /uva10344.in "," r ", stdin); # endif while (scanf (" % d ", A, A + 1, A + 2, A + 3, A + 4) = 5 & (A [0] | A [1] | A [2] | A [3] | A [4]) {Possible = 0; DFS (0, 0); If (possible) puts ("possible"); else puts ("impossible");} return 0 ;}

Return immediately after finding, time 0.945

Learning point: when DFS is used to determine the return value of the extended point, if it has been successful, it will be returned directly.

# Include <cstdio> # include <cstring> # include <iostream> # include <string> # include <algorithm> using namespace STD; int A [5]; int vis [5]; bool DFS (int d, int s) {If (D = 5) return S = 23; For (INT I = 0; I <5; I ++) if (! Vis [I]) {vis [I] = 1; // The first number if (D = 0) {If (DFS (D + 1, a [I]) return true;} else {If (DFS (D + 1, S + A [I]) return true; If (DFS (D + 1, s-A [I]) return true; If (DFS (D + 1, S * A [I]) return true;} vis [I] = 0 ;} return false;} int main () {# ifndef online_judge freopen (". /uva10344.in "," r ", stdin); # endif while (scanf (" % d ", A, A + 1, A + 2, A + 3, A + 4) = 5 & (A [0] | A [1] | A [2] | A [3] | A [4]) {memset (VIS, 0, sizeof (VIS); If (DFS (0, 0) puts ("possible"); else puts ("impossible ");} return 0 ;}

 

Next_permutation calculation almost times out: 2.388

#include<cstdio>#include<cstring>#include<iostream>#include<string>#include<algorithm>using namespace std;int a[5];bool check(int k){    int s=a[0];    for(int j=1;j<5;j++)    {        switch(k%3)        {            case 0:                s+=a[j]; break;            case 1:                s-=a[j]; break;            case 2:                s*=a[j]; break;        }        k/=3;    }    return s==23;}int main(){#ifndef ONLINE_JUDGE    freopen("./uva10344.in", "r", stdin);#endif    while(scanf("%d%d%d%d%d", a, a+1, a+2, a+3, a+4)==5             && (a[0] || a[1] || a[2] || a[3] || a[4]))    {        sort(a, a+5);        int ok=0;        do        {            for(int i=0;i<81;i++) if(check(i))            {                ok=1;                break;            }        }while(!ok && next_permutation(a, a+5));        if(ok)            puts("Possible");        else            puts("Impossible");    }    return 0;}