HDU3344(小广搜+小暴力

Kakuro Extension Extension

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 507    Accepted Submission(s): 248

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 028XXX 017XXX 028XXX XXXXXXX XXXXXXX 02222 ....... ....... ....... 010XXX XXX34 ....... ....... ....... ....... ....... XXX14 ....... ....... 01613 ....... ....... XXX22 ....... ....... ....... ....... XXXXXXX XXXXXXX XXX16 ....... ....... XXXXXXX XXXXXXX XXXXXXX 001XXX 020XXX 027XXX 021XXX 028XXX 014XXX 024XXX XXX35 ....... ....... ....... ....... ....... ....... ....... XXXXXXX 00734 ....... ....... ....... ....... ....... ....... XXX43 ....... ....... ....... ....... ....... ....... ....... XXX30 ....... ....... ....... ....... ....... ....... XXXXXXX

 

 

Author

shǎ崽

 

 

Source

HDU2010省赛集训队选拔赛（校内赛）

 

 

Recommend

 

 

 

 

 

#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
using namespace std;
char a[105][105];

void pd(int x,int y,int m,int n)
{
    if(a[x][y]>='1'&&a[x][y]<='9')
    {
        printf(".......");
        return ;
    }
    int sum1=0,sum2=0;
    int flag1=0;
    int flag2=0;
    if(x==m)
        flag1=1;
    if(y==n)
        flag2=1;
    for(int i=x+1; i<=m; i++)
    {
        if(a[x+1][y]=='_')
            flag1=1;
        if(a[i][y]=='_')
            break;
        sum1=sum1+(a[i][y]-'0');
    }
    for(int j=y+1; j<=n; j++)
    {
        if(a[x][y+1]=='_')
            flag2=1;
        if(a[x][j]=='_')
            break;
        sum2=sum2+(a[x][j]-'0');
    }
    if(flag1==1&&flag2==1)
    {
        printf("XXXXXXX");

    }
    else if(flag1==1&&flag2==0)
    {
            printf("XXX");
            printf("\");
            printf("%03d",sum2);

    }
    else if(flag1==0&&flag2==1)
    {
            printf("%03d",sum1);
            printf("\");
            printf("XXX");
    }
    else if(flag1==0&&flag2==0)
    {
            printf("%03d\%03d",sum1,sum2);
    }

}
int main()
{
    int m,n;
    int t;
    scanf("%d",&t);
    while(t--)
    {
        memset(a,0,sizeof(a));
        scanf("%d%d",&m,&n);
        getchar();
        for(int i=1; i<=m; i++)
        {
            for(int j=1; j<=n; j++)
            {
                //scanf("%c",&a[i][j]);
                cin >> a[i][j];
            }

        }
        char ch=' ';
        for(int i=1; i<=m; i++)
        {
            for(int j=1; j<=n; j++)
            {
                /*if(a[1][1]!='_')
                    a[1][1]='_';*/
                pd(i,j,m,n);
                if(j!=n)
                    printf("%c",ch);
            }
            printf(" ");
        }
        printf(" ");
    }
    return 0;
}