H题；
In this problem you need to make a triangle ,just like the sample out. The element in the ith row and jth 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<=20) in a line which mentioned above.
Output
For each test case, print out the triangle. the triangle separated by a blank line.
If the product is more than 9 (product > 9) you should print a space, or you should print two space.The final number of each line don’t need print space.
Sample Input
3
5
6
7
Sample Output
1 2 3 4 5
2 4 6 8
3 6 9
4 8
5

1 2 3 4 5 6
2 4 6 8 10
3 6 9 12
4 8 12
5 10
6

1 2 3 4 5 6 7
2 4 6 8 10 12
3 6 9 12 15
4 8 12 16
5 10 15
6 12
7

题意: 打印三角形，ij的值的打印；
解题思路： 输出数字按格式输出即可，ij，那么就两层for循环，输出时,注意判断数字是否小于九，是则后面一个空格，否则要有两个空格；那么输出的数字规律是：i*j;
特别要注意的是打印的格式；
乘积小于9每个数之见按一个空格，大于等于九就 是两个空格；

#include<cstdio>
#include<cstring>
#include<cmath
#include<algorithm>
#include<iostream>
using namespace std;
int main()
{
    int n,m,t=0;
    scanf("%d",&n);
    while(n--)
    {
        scanf("%d",&m);
        int s=m;
        for(int i=1;i<=m;i++)
        {
            for(int j=1;j<=s;j++)
            {
                printf("%d",i*j);
                if(i*j>9&&j!=s)
                    printf(" ");
                else if(i*j<=9&&j!=s)
                    printf("  ");//小于九则打印两个空格；
            }
            printf("\n");
            s--;
        }
        if(n)
            printf("\n");
    }
    return 0;
}