Kakuro Extension Extension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)

Problem Description

You know ,I‘m a lazy guy and write problem description is a very very boring thing.So , I would not repeat the rule of Kakuro again , Please look at this.But
things are different again,contray to the problem above,this time you should work out the input file according to the output file.

Input

The first line of the inputs is T, which stands for the number of test cases you need to solve.

Then T case follow:

Each test case starts with a line contains two numbers N,M (2<=N,M<=100)and then N lines follow, each line contains M columns, either ‘_’ or 1~9. You can assume that the first column of the first line is ’_’.

Output

Output N lines, each line contains M parts, each part contains 7 letters. The m parts are seperated by spaces.Output a blank line after each case.

Sample Input

2
6 6
_ _ _ _ _ _
_ _ 5 8 9 _
_ 7 6 9 8 4
_ 6 8 _ 7 6
_ 9 2 7 4 _
_ _ 7 9 _ _
5 8
_ _ _ _ _ _ _ _
_ 1 9 9 1 1 8 6
_ _ 1 7 7 9 1 9
_ 1 3 9 9 9 3 9
_ 6 7 2 4 9 2 _

Sample Output

XXXXXXX XXXXXXX 028\XXX 017\XXX 028\XXX XXXXXXX
XXXXXXX 022\022 ....... ....... ....... 010\XXX
XXX\034 ....... ....... ....... ....... .......
XXX\014 ....... ....... 016\013 ....... .......
XXX\022 ....... ....... ....... ....... XXXXXXX
XXXXXXX XXX\016 ....... ....... XXXXXXX XXXXXXX

XXXXXXX 001\XXX 020\XXX 027\XXX 021\XXX 028\XXX 014\XXX 024\XXX
XXX\035 ....... ....... ....... ....... ....... ....... .......
XXXXXXX 007\034 ....... ....... ....... ....... ....... .......
XXX\043 ....... ....... ....... ....... ....... ....... .......
XXX\030 ....... ....... ....... ....... ....... ....... XXXXXXX

#include<cstdio>
#include<algorithm>
#include<iostream>
#include<cstring>
#include<cmath>

using namespace std;

struct node {
    int r;
    int d;
    bool ok;
} a[111][111];

int b[111][111];

int n,m;
int num(int x) {
    if(x==0)return 1;
    int ans=0;
    while(x)
        ans++,x/=10;
    return 3-ans;
}

int main() {
    int t;
    cin>>t;
    while(t--) {
        char c;
        scanf("%d%d",&n,&m);
        memset(a,0,sizeof a);
        memset(b,0,sizeof b);
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=m; j++) {
                scanf(" %c",&c);
                if(c=='_') {
                    a[i][j].ok=1;
                } else {
                    b[i][j]=c-'0';
                }
            }
        }
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=m; j++) {

                if(a[i][j].ok) {
                    for(int k=j+1; k<=m; k++) {
                        if(a[i][k].ok)break;
                        a[i][j].r+=b[i][k];
                    }
                    for(int k=i+1; k<=n; k++) {
                        if(a[k][j].ok)break;
                        a[i][j].d+=b[k][j];
                    }
                }
            }
        }
        for(int i=1; i<=n; i++) {
            for(int j=1; j<=m; j++) {
                if(a[i][j].ok) {
                    if(a[i][j].d) {
                        int h=num(a[i][j].d);
                        for(int i=0; i<h; i++)
                            printf("0");
                        printf("%d",a[i][j].d);
                    } else
                        printf("XXX");
                    if(a[i][j].r==0&&a[i][j].d==0) {
                        printf("XXXX");
                    } else if(a[i][j].d&&a[i][j].r==0)
                        printf("\\XXX");
                    else{
                              printf("\\");
                        int h=num(a[i][j].r);
                        for(int i=0; i<h; i++)
                            printf("0");
                        printf("%d",a[i][j].r);
                    }
                } else if(b[i][j])
                    printf(".......");
                if(j!=m)printf(" ");
                else    printf("\n");
            }
        }
          printf("\n");
    }
    return 0;
}