Baby Ming and Matrix games
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/65536 K (Java/Others)
Total Submission(s): 1150 Accepted Submission(s): 298
Problem Description
These few days, Baby Ming is addicted to playing a matrix game.
Given a n∗m
matrix, the character in the matrix(i∗2,j∗2) (i,j=0,1,2...)
are the numbers between 0−9.
There are an arithmetic sign (‘+’, ‘-‘, ‘∗’,
‘/’) between every two adjacent numbers, other places in the matrix fill with ‘#’.
The question is whether you can find an expressions from the matrix, in order to make the result of the expressions equal to the given integer
sum.
(Expressions are calculated according to the order from left to right)
Get expressions by the following way: select a number as a starting point, and then selecting an adjacent digital X to make the expressions, and then, selecting the location of X for the next starting point. (The number in same place can’t be used twice.)
Given a n∗m
matrix, the character in the matrix(i∗2,j∗2) (i,j=0,1,2...)
are the numbers between 0−9
.There are an arithmetic sign (‘+’, ‘-‘, ‘∗
’,‘/’) between every two adjacent numbers, other places in the matrix fill with ‘#’.
The question is whether you can find an expressions from the matrix, in order to make the result of the expressions equal to the given integer
sum
.(Expressions are calculated according to the order from left to right)
Get expressions by the following way: select a number as a starting point, and then selecting an adjacent digital X to make the expressions, and then, selecting the location of X for the next starting point. (The number in same place can’t be used twice.)
Input
In the first line contains a single positive integer
T,
indicating number of test case.
In the second line there are two odd numbers n,m,
and an integer sum(−1018<sum<1018,
divisor 0 is not legitimate, division rules see example)
In the next n
lines, each line input m
characters, indicating the matrix. (The number of numbers in the matrix is less than
15)
1≤T≤1000
T
,indicating number of test case.
In the second line there are two odd numbers n,m
,and an integer sum(−10
18<sum<1018,divisor 0 is not legitimate, division rules see example)
In the next n
lines, each line input m
characters, indicating the matrix. (The number of numbers in the matrix is less than
15
)1≤T≤1000
Output
Print Possible if it is possible to find such an expressions.
Print Impossible if it is impossible to find such an expressions.
Print Impossible if it is impossible to find such an expressions.
Sample Input
3
3 3 24
1*1
+#*
2*8
1 1 1
1
3 3 3
1*0
/#*
2*6
Sample Output
Possible
Possible
PossibleHintThe first sample：1+2*8=24
The third sample：1/2*6=3
Source
Recommend
传递的参数都是整数，如果遇到比较大的分母可能表达式的值为0，所以传递的b表示分母，遇到除法就把数乘到分母上，加法或者减法就乘以分母然后加减到分子上，最后判断的时候看是否等于sum*b
#include<cstdio>
#include<queue>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
int dx[4]={2,-2,0,0};
int dy[4]={0,0,2,-2};
char str[50][50];
bool flag;
int vis[50][50],n,m;
__int64 sum;
bool judge(int x,int y)
{
return x>=0&&x<n&&y>=0&&y<m;
}
void dfs(int x,int y,__int64 a,__int64 b)
{
if(flag) return ;
if(a==b*sum)
{
flag=true;
return;
}
for(int i=0;i<4;i++)
{
int xx=x+dx[i];
int yy=y+dy[i];
if(!judge(xx,yy)||vis[xx][yy]||str[xx][yy]<'0'||str[xx][yy]>'9'||str[xx][yy]=='#')
continue;
__int64 v=str[xx][yy]-'0';
int mx=(x+xx)>>1;
int my=(y+yy)>>1;
__int64 aa=a,bb=b;
if(str[mx][my]=='/'&&v==0) continue;
vis[xx][yy]=1;
if(str[mx][my]=='*'){aa*=v;}
else if(str[mx][my]=='/'){bb*=v;}
else if(str[mx][my]=='+') {aa+=bb*v;}
else {aa-=bb*v;}
dfs(xx,yy,aa,bb);
vis[xx][yy]=0;
}
}
int main()
{
int t;
cin>>t;
while(t--)
{
scanf("%d%d",&n,&m);
scanf("%I64d",&sum);
flag=false;
memset(str,'\0',sizeof(str));
for(int i=0;i<n;i++)
scanf("%s",str[i]);
for(int i=0;i<n;i++)
{
for(int j=0;j<m;j++)
{
if(str[i][j]<='9'&&str[i][j]>='0')
{
memset(vis,0,sizeof(vis));
vis[i][j]=1;
__int64 val=str[i][j]-'0';
dfs(i,j,val,1);
if(flag) break;
}
}
if(flag) break;
}
printf(flag?"Possible\n":"Impossible\n");
}
return 0;
}