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;
else
cout << "Impossible" << endl;
}else
break;
}
// printf("%.2lf\n", (double)clock()/CLOCKS_PER_SEC);
return 0;
}
全排列+DFS