An easy problem

Time Limit: 5000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 10669    Accepted Submission(s): 7288

Problem Description

In this problem you need to make a multiply table of N * N ,just like the sample out. The element in the i^th row and j^th column should be the product(乘积) of i and j.

Input

The first line of input is an integer C which indicate the number of test cases.

Then C test cases follow.Each test case contains an integer N (1<=N<=9) in a line which mentioned above.

Output

For each test case, print out the multiply table.

Sample Input

2
1
4

Sample Output

1
1 2 3 4
2 4 6 8
3 6 9 12
4 8 12 16

Hint
There is no blank space at the end of each line.

一个简单的问题

问题描述

在这个问题中，你需要做一个N * N的乘法表，就像样本一样。i行和第j列中的元素应该是产品(乘积)i和j。

输入

输入的第一行是一个整数C，表示测试用例的数量。

然后是C测试用例。每个测试用例在上面提到的一行中包含一个整数N (1<=N<=9)。

输出

对于每个测试用例，打印乘法表。

样例输入

2

1

4

样例输出

1

1 2 3 4

2 4 6 8

3 6 9 12

4 8 12 16

提示

每行的末尾没有空格。

代码：

#include<stdio.h>
int main()
{
    int i,t,n,j,k,f;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(i=1,j=1;i<=n;i++,j++)
        {
            f=i;
            for(k=1;k<=n;i+=j,k++)
	        if(k==n)
		    printf("%d",i);
	        else
                    printf("%d ",i);
                i=f;
            printf("\n");
        }
    }
    return 0;
}