目录
一、题目
二、题目分析
三、
一、题目
传送门
C. No Prime Differences
time limit per test
2 seconds
memory limit per test
256 megabytes
input
standard input
output
standard output
You are given integers n and m. Fill an n by m grid with the integers 11 through n⋅m, in such a way that for any two adjacent cells in the grid, the absolute difference of the values in those cells is not a prime number. Two cells in the grid are considered adjacent if they share a side.
It can be shown that under the given constraints, there is always a solution.
Input
The first line of the input contains a single integer t (1≤t≤1000) — the number of test cases. The description of the test cases follows.
The first and only line of each test case contains two integers n and m (4≤n,m≤1000) — the dimensions of the grid.
It is guaranteed that the sum of n⋅m over all test cases does not exceed 106106.
Output
For each test case, output n lines of m integers each, representing the final grid. Every number from 1 to n⋅mshould appear exactly once in the grid.
The extra spaces and blank lines in the sample output below are only present to make the output easier to read, and are not required.
If there are multiple solutions, print any of them.
Example
input
3
4 4
5 7
6 4
output
16 7 1 9 12 8 2 3 13 4 10 11 14 5 6 15 29 23 17 9 5 6 2 33 27 21 15 11 7 1 32 31 25 19 20 16 10 26 30 24 18 14 8 4 35 34 28 22 13 12 3 2 3 7 11 8 9 1 10 17 13 5 4 18 14 6 12 19 23 15 21 20 24 16 22
Note
The first sample case corresponds to the picture above. The only absolute differences between adjacent elements in this grid are 11, 44, 66, 88, and 99, none of which are prime.
二、题目分析
题目n*m表格要求每一个数的上下左右之差均不为素数，画出两个图之后我们发现，如果按照正常顺序写的话，这个数的左右之差为1符合题意， 上下之差为m，这样的话上下之差可能符合也可能不符合;所以我们按照m是否是素数来讨论：
1.m不为素数，直接按原顺序输出即可
2.m为素数，我们知道素数的倍数一定不是素数，所以我们考虑将原顺序中的行顺序都换一下，这样左右之差还是1，上下之差为素数的倍数。
至于怎么样的换，我们不妨先输出奇数行，再输出偶数行，但是有一个点我们需要注意，当n等于4的时候，这样我们输出顺序就变为了 行顺序:1 3 2 4，这样 3 2之差还是素数，所以我们还是先输出奇数再输出偶数，但是先从大的开始输出，也即行顺序： 3 1 4 2这样即可保证答案正确
三、
#include<bits/stdc++.h>
using namespace std;
bool isprime(int x)
{
if(x<2)return false;
for(int i=2;i<=x/i;i++)
{
if(x%i==0)return false;
}
return true;
}
int main()
{
int t;cin>>t;
while(t--)
{
int x,y;cin>>x>>y;
if(isprime(y))
{
for(int i=x-1;i>=1;i-=2)
{
for(int j=1;j<=y;j++)
{
cout<<y*(i-1)+j<<" ";
}cout<<"\n";
}
for(int i=x;i>=1;i-=2)
{
for(int j=1;j<=y;j++)
{
cout<<y*(i-1)+j<<" ";
}cout<<"\n";
}
}
else
{
for(int i=1;i<=x*y;i++)
{
cout<<i<<" ";
if(i%y==0)cout<<"\n";
}
}
cout<<endl;
}
}
