| Time Limit: 2000MS | Memory Limit: 65536KB | 64bit IO Format: %lld & %llu |
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
Source
Asia 2000, Tehran (Iran)
分析:
非常简单的找规律题。
ac代码:
#include <iostream>
#include<cstdio>
using namespace std;
int main()
{
int n,x,y;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&x,&y);
if(x==y&&x%2==0)
printf("%d\n",x+y);
else if(x==y&&x%2==1)
printf("%d\n",x+y-1);
else if(x-y==2&&x%2==1)
printf("%d\n",x+y-1);
else if(x-y==2&&x%2==0)
printf("%d\n",x+y);
else
printf("No Number\n");
}
return 0;
}
#include<cstdio>
using namespace std;
int main()
{
int n,x,y;
scanf("%d",&n);
while(n--)
{
scanf("%d%d",&x,&y);
if(x==y&&x%2==0)
printf("%d\n",x+y);
else if(x==y&&x%2==1)
printf("%d\n",x+y-1);
else if(x-y==2&&x%2==1)
printf("%d\n",x+y-1);
else if(x-y==2&&x%2==0)
printf("%d\n",x+y);
else
printf("No Number\n");
}
return 0;
}
这篇博客探讨了一种在平面上按照特定规律书写数字的方法,并提供了一个程序来读取坐标并输出对应的数字或提示未找到数字。
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