UVa 10344 - 23 out of 5 递归回溯

Problem I

23 Out of 5

Input: standard input

Output: standard output

Time Limit: 1 second

Memory Limit: 32 MB

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 should not be processed.

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

---

Thomas Strohmann

译文：

你的任務是寫一個程式，看看是否能在5個數字間插入一些運算子使得結果為23。

考慮以下的運算式結果是否可能等於23。

(((a₁ O₁ a₂) O₂ a₃) O₃ a₄) O₄ a₅

在這裡a₁～a₅為5個給你的整數（順序可以隨便排列，但一定都要出現一次），O₁～O₄為運算子，內容為{+,-,*}其中一個。如果你還不清楚的話，以下面的例子來說明：

輸入5個整數2,3,5,711
你可以找到有一組運算式 (((11*3)-5)+2)-7=23，所以輸出Possible。（當然，可以得到23的答案的運算式可能不只一組）

若輸入的5個整數為1,1,1,1,1
那你就找不到任一種運算式的組合可以使答案為23。所以輸出Impossible。

Input

每一測試資料一列，有5個整數。每個整數均介於0到50之間。當輸入為5個0時代表輸入結束。測試資料總共不會超過25列。

Output

根據輸入的5個整數，判斷是否可能找到使其答案為23的運算式。請參考Sample Output

#include <iostream>#include <cstring>#include <cstdio>#include <ctime>#include <algorithm>using namespace std;int num[5];int vis[5];char flag2[3]={'+','-','*'};int vis2[3];int temp[5];char calSignal[4];bool ans=false;void init(){ans = false;memset(num, 0, sizeof(num));memset(vis, 0, sizeof(vis));memset(vis2, 0, sizeof(vis2));}bool cal(){int value=0;for(int i=0; i < 4; i++){switch(calSignal[i]){case '+': {if(i==0)value = temp[i]+temp[i+1];else value += temp[i+1];break;}case '-': {if(i==0)value = temp[i]-temp[i+1];else value -= temp[i+1];break;}case '*':{if(i==0)value = temp[i]*temp[i+1];else value *= temp[i+1];break;}}}if(value==23 || value==-23)return true;return false;}// 同时遍历两种可能性 TLE void f(int number, int signal){if(ans) return ;if(number==4 && signal==3){if(cal())ans = true;return ;}for(int i=0; i < 5; i++){if(!vis[i]){vis[i]=1;temp[number] = num[i];f(number+1, signal);vis[i]=0;}}for(int i=0; i < 3; i++){if(!vis2[i]){vis2[i]=1;calSignal[signal] = flag2[i];f(number, signal+1);vis2[i] = 0;}}}void dfs(int length, int sum){if(length==5){if(sum==23)ans = true;return ;}sum += num[length];dfs(length+1, sum);sum -= num[length];if(ans) return;sum -= num[length];dfs(length+1, sum);sum += num[length];if(ans) return;sum *= num[length];dfs(length+1, sum);sum /= num[length];if(ans) return ;}void solve(){sort(num, num+5);do{dfs(1, num[0]);if(ans)return ;}while(next_permutation(num, num+5));}bool read(){for(int i=0; i < 5; i++){cin >> num[i];if(num[i]==0) return false;}return true;}int main(){//freopen("in.txt","r",stdin);while(1){init();bool flag = read();if(flag){solve(); if(ans)cout << "Possible" << endl;elsecout << "Impossible" << endl;}elsebreak;}//printf("%.2lf\n", (double)clock()/CLOCKS_PER_SEC);return 0;} 

全排列+DFS