这里有道题，大侠们给点思路好嘛！
利用队列写出打印杨辉三角形的前n行的程序。（杨辉三角形的特点是两个腰上的数字都为1，其它位置上的数字是其上一行中与之相邻的两个整数之和。）
各位大侠给点思路好嘛！

[解决办法]
http://topic.csdn.net/t/20021109/08/1161618.html
[解决办法]

C/C++ code

#include <stdio.h>int main(){    int n,i,k,j = 0,matrix[12][12];//这里最多只容纳12行的杨辉三角    printf("Enter n:");scanf("%d",&n);    k = n - 1;    for (i = 0;i < n;i ++)    {        for (j = 0;j < k;j ++) printf("%3s"," ");        matrix[i][0] = 1;printf("%3d%3s",matrix[i][0]," ");        for (j = 1;j < i;j ++)        {            matrix[i][j] = matrix[i-1][j-1] + matrix[i-1][j];            printf("%3d%3s",matrix[i][j]," ");        }        if(i == 0)        {             printf("\n");        } else { matrix[i][j] = 1;printf("%3d%3s\n",matrix[i][j]," "); }        k --;    }    return 1;}
[解决办法]
#include <stdio.h>
int main()
{
   int n,i,k,j = 0,matrix[12][12];//这里最多只容纳12行的杨辉三角
   printf("Enter n:");scanf("%d",&n);
   k = n - 1;
   for (i = 0;i < n;i ++)
   {
       for (j = 0;j < k;j ++) printf("%3s"," ");
       matrix[i][0] = 1;printf("%3d%3s",matrix[i][0]," ");
       for (j = 1;j < i;j ++)
       {
           matrix[i][j] = matrix[i-1][j-1] + matrix[i-1][j];
           printf("%3d%3s",matrix[i][j]," ");
       }
       if(i == 0)
       {  
           printf("\n");
       } else { matrix[i][j] = 1;printf("%3d%3s\n",matrix[i][j]," "); }
       k --;
   }
   return 1;
}
C/C++ code
#include <stdio.h>int main(){    int n,i,k,j = 0,matrix[12][12];//这里最多只容纳12行的杨辉三角    printf("Enter n:");scanf("%d",&n);    k = n - 1;    for (i = 0;i < n;i ++)    {        for (j = 0;j < k;j ++) printf("%3s"," ");        matrix[i][0] = 1;printf("%3d%3s",matrix[i][0]," ");        for (j = 1;j < i;j ++)        {            matrix[i][j] = matrix[i-1][j-1] + matrix[i-1][j];            printf("%3d%3s",matrix[i][j]," ");        }        if(i == 0)        {             printf("\n");        } else { matrix[i][j] = 1;printf("%3d%3s\n",matrix[i][j]," "); }        k --;    }    return 1;}
[解决办法]
#include<stdio.h>
define n 10   要打印几行可以改N的数值。
main()
  {int i,j,a[n][n] ;
 for(i=0;i<=9;i++)    将数组第一列和对角线置0
  for(i=0;i<=9;i++)
   a[i][j]=a[0][j]=1;
 for(i=1;i<=9;i++)
  for(i=1;i<=9;i++)
    a[i][j]=a[i-1][j-1]+a[i-1][j];
 for(i=1;i<=9;i++)
   for(i=1;i<=9;i++)
 prinft("%d",a[i][j]);
}
思路是这样的，、
先将数组第一列和对角线置0；
然后下中间的数就等于前一行同列和前一行的前一列之和
1
11
121   如出2=1+1
1331
13641
输出时注意一下格式就行了。