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最新推荐文章于 2025-08-17 21:38:20 发布

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原文链接：http://www.cnblogs.com/13224ACMer/p/4445093.html

文章标签：

#java

本文介绍了一种解决Kakuro数字谜题的方法，并提供了一个具体的实现案例。该程序能够根据给定的输出文件反向生成输入文件内容，通过解析数字布局来形成特定的输出格式。

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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 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
  

Author

shǎ崽

Source

HDU2010省赛集训队选拔赛（校内赛）

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#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("\n");
        }
        printf("\n");
    }
    return 0;
}