杨辉三角

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 67572    Accepted Submission(s): 27943

Problem Description

还记得中学时候学过的杨辉三角吗？具体的定义这里不再描述，你可以参考以下的图形：
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1

 

Input

输入数据包含多个测试实例，每个测试实例的输入只包含一个正整数n（1<=n<=30），表示将要输出的杨辉三角的层数。

 

Output

对应于每一个输入，请输出相应层数的杨辉三角，每一层的整数之间用一个空格隔开，每一个杨辉三角后面加一个空行。

 

Sample Input

2 3

 

Sample Output

11 111 11 2 1

 

Author

lcy

 

Source

C语言程序设计练习（五）

 

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#include<iostream>#include<cstdlib>#include<cstdio>#include<cmath>#include<cstring>#include<string>#include<cstdlib>#include <set>#include<algorithm>#include <limits.h>#include <ctype.h>#include <map>#include <stack>#include <sstream>typedef long long LL;using namespace std;const int N = 105;int data[N][N];int dp[N][N];char c[100][100] = {' '};#define PI 3.1415926void fun() {    data[0][0] = 1;    data[1][0] = 1;    data[1][1] = 1;    for(int i = 2; i <= 30; i++) {        for(int j = 0; j <= i; j++) {            if(j == 0 || j == i) data[i][j] = 1;            else data[i][j] = data[i - 1][j - 1] + data[i - 1][j];        }    }//    for(int i = 0; i <= 30; i++){//        for(int j = 0; j <= i; j++){//            cout << data[i][j] << " ";//        }//        cout << endl;//    }}int main() {    //freopen("in.txt","r",stdin);    //freopen("out.txt","w",stdout);    int n;    fun();    int flag = 0;    while(scanf("%d", &n) != EOF) {        //if(flag) cout << endl;        for(int i = 0; i < n; i++) {            for(int j = 0; j <= i; j++) {                if(j == 0) cout << data[i][j] ;                else cout << " "<< data[i][j] ;            }            cout << endl;        }        cout << endl;        flag = 1;    }    return 0;}