hdu 1391 Number Steps

最新推荐文章于 2020-01-11 20:46:59 发布

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本文介绍了一种在特定平面坐标系中确定整数位置的方法。通过观察坐标系中的数字排列规律，给出了一个简单有效的算法来找出指定坐标点上的数字。输入为坐标点(x, y)，输出该点上对应的数字或提示未定义。

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Number Steps

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 5682 Accepted Submission(s): 3482

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

思路：其实是找规律，仔细观察，当x是偶数时，x上面的两个数等于x+y，当x是奇数时，x上面的两个数等于x+y-1.
AC代码：

#include<iostream>
#include<cstdio>
using namespace std;
int main(){
    int n;
    scanf("%d", &n);
    while(n--){
        int x, y;
        scanf("%d %d", &x, &y);
        if(x == y || x == y+2){
            if(x % 2 == 0 ) 
                printf("%d\n", x + y);
            else 
                printf("%d\n", x + y -1);   

        }
        else
            printf("No Number\n");
    }
    return 0;
}