题解 | #杨辉三角#

#include<stdio.h>
int main(){
    int n = 0;
    scanf("%d", &n);
    int** t;
    t = (int**)malloc(sizeof(int*)* n);
    for (int i = 0; i < n; i++) {
         t[i] = (int*)malloc(sizeof(int)* n);
    }
        //创建初始矩阵(n=4)
        //1 0 0 0
        //1 0 0 0
        //1 0 0 0
        //1 0 0 0
    for (int j = 0; j < n; j++) {           
        t[j][0] = 1;//第一列都初始化为1
        for (int k = 1; k < n; k++) {//其余初始化为0
            t[j][k] = 0;
        }
    }
    for (int i = 1; i < n; i++) {//计算除第一列第一行以外的数
        for (int j = 1; j < n; j++) {
            t[i][j] = t[i - 1][j - 1] + t[i - 1][j];//当前数等于上一行数字与上一行数字前一位数字之和
        }
    }
    for (int i = 0; i < n; i++) {
        for (int j = 0; j < n; j++) {
            if (t[i][j]) {//不为0即打印
                printf("%5d", t[i][j]);
            }
        }
        printf("\n");
    }
        for(int i=0;i<n;i++)   
           free(t[i]);    
           free(t);
 
  

    return 0;
}