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。

import java.util.Scanner;

public class Main{
 static int[][] db = new int[5005][2];
 public static void main(String[] args) {
 dabiao();
 Scanner sc = new Scanner(System.in);
 int t = sc.nextInt();
 while(t-->0){
 int x = sc.nextInt();
 int y = sc.nextInt();
 if(x==0&&y==0){
 System.out.println(0);
 continue;
 }

 if(x==1&&y!=1){
 System.out.println("No Number");
 continue;
 }

 if(x==1&&y==1){
 System.out.println(1);
 continue;
 }

 if(x==y||x==(y+2)){
 if(x==(y+2)){
 System.out.println(db[x][0]);
 }else{
 System.out.println(db[x][1]);
 }
 }else{
 System.out.println("No Number");
 }
 }
 }

 private static void dabiao() {
 int num=2;
 db[0][0]=0;
 boolean is = true;
 for(int i=2;i<=5003;i++){
 if(is){
 db[i][0]=num;
 num++;
 db[i+1][0]=num;
 num++;
 is=!is;
 }else if(!is){
 db[i-1][1]=num;
 num++;
 db[i][1]=num;
 num++;
 is=!is;
 }
 }
// System.out.println(db[2][0]);
// System.out.println(db[2][1]);
// System.out.println(db[3][0]);
// System.out.println(db[3][1]);
// System.out.println(db[4][0]);
// System.out.println(db[4][1]);
// System.out.println(db[5][0]);
// System.out.println(db[5][1]);
 }
}

  

 

  
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文章来源: chenhx.blog.csdn.net，作者：谙忆，版权归原作者所有，如需转载，请联系作者。

原文链接：chenhx.blog.csdn.net/article/details/51105667