HDOJ 1391 Number Steps(打表DP)

Problem Description
Starting from point (0,0) on a plane, we have written all non-negative integers 0, 1, 2,… as shown in the figure. For example, 1, 2, and 3 has been written at points (1,1), (2,0), and (3, 1) respectively and this pattern has continued.

You are to write a program that reads the coordinates of a point (x, y), and writes the number (if any) that has been written at that point. (x, y) coordinates in the input are in the range 0…5000.

Input
The first line of the input is N, the number of test cases for this problem. In each of the N following lines, there is x, and y representing the coordinates (x, y) of a point.

Output
For each point in the input, write the number written at that point or write No Number if there is none.

Sample Input
3
4 2
6 6
3 4

Sample Output
6
12
No Number

不能直接开[5005][5005]的数组，这样内存不够。
因为大部分数据没用，没列只有2个有效数据，所以开[5005][2]就可以了。

1import java.util.Scanner;
2
3public class Main{
4    static int[][] db = new int[5005][2];
5    public static void main(String[] args) {
6   
7   
8   
9        dabiao();
10        Scanner sc = new Scanner(System.in);
11        int t = sc.nextInt();
12        while(t-->0){
13   
14   
15   
16            int x = sc.nextInt();
17            int y = sc.nextInt();
18            if(x==0&&y==0){
19   
20   
21   
22                System.out.println(0);
23                continue;
24            }
25
26            if(x==1&&y!=1){
27   
28   
29   
30                System.out.println("No Number");
31                continue;
32            }
33
34            if(x==1&&y==1){
35   
36   
37   
38                System.out.println(1);
39                continue;
40            }
41
42            if(x==y||x==(y+2)){
43   
44   
45   
46                if(x==(y+2)){
47   
48   
49   
50                    System.out.println(db[x][0]);
51                }else{
52   
53   
54   
55                    System.out.println(db[x][1]);
56                }
57            }else{
58   
59   
60   
61                System.out.println("No Number");
62            }
63        }
64    }
65
66    private static void dabiao() {
67   
68   
69   
70        int num=2;
71        db[0][0]=0;
72        boolean is = true;
73        for(int i=2;i<=5003;i++){
74   
75   
76   
77            if(is){
78   
79   
80   
81                db[i][0]=num;
82                num++;
83                db[i+1][0]=num;
84                num++;
85                is=!is;
86            }else if(!is){
87   
88   
89   
90                db[i-1][1]=num;
91                num++;
92                db[i][1]=num;
93                num++;
94                is=!is;
95            }
96        }
97//      System.out.println(db[2][0]);
98//      System.out.println(db[2][1]);
99//      System.out.println(db[3][0]);
100//      System.out.println(db[3][1]);
101//      System.out.println(db[4][0]);
102//      System.out.println(db[4][1]);
103//      System.out.println(db[5][0]);
104//      System.out.println(db[5][1]);
105    }
106}

本文同步分享在 博客“谙忆”（CSDN）。
如有侵权，请联系 support@oschina.cn 删除。
本文参与“OSC源创计划”，欢迎正在阅读的你也加入，一起分享。