杭电OJ题 1391 Number Steps 解题报告

Number Steps

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 2697    Accepted Submission(s): 1656

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

 
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主要是找规律。当x==y时 当x是偶数时，为x*2，如果是奇数时，为2*n-1，其他的数的话如果y < x，则为y==x时的数减去2就行了

/***********************
*   程序名:Number Steps.c
*   功能:ACM
************************/

#include <stdio.h>

int main()
{
    int N, x, y;
    scanf("%d", &N);
    while(N--) {
        scanf("%d %d", &x, &y);
        if(x == y || x - y == 2) {
            if(x == y) {
                if(x % 2 == 0) {
                    printf("%d\n", x*2);
                }
                else {
                    printf("%d\n", x*2-1);
                }
            }
            else {
                if(x % 2 == 0) {
                    printf("%d\n", x*2-2);
                }
                else {
                    printf("%d\n", x*2-3);
                }
            }
        }
        else {
            printf("No Number\n");
        }
    }
    return 0;
}