1 solutions
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0
C :
int main(){ int n,i,j,z; scanf("%d",&n); //控制输出的行数 for(i = 1;i <= n;i++){ //控制每行输出空格量 for(j = 0;j < n - i;j++){ printf(" "); } //控住输出的* for(z=1;z <= 2 * i + 1;z++){ printf("*"); } //负责换行 printf("\n"); } }
C++ :
#include<iostream> using namespace std; int main(){ int i,j,k,n; cin>>n; for(i = 1;i <= n;i++) { for(j = 1;j <= n - i;j++) { cout<<" "; } for(k = 1;k <= 2 * i + 1;k++) { cout<<"*"; } cout<<endl; } return 0; }
Python :
n = int(input()); for i in range(1, n + 1): for k in range(1, n -i + 1): print(' ',end = '') for x in range(1, 2 * i +2): print('*', end = '') print()
- 1
Information
- ID
- 10879
- Time
- 1000ms
- Memory
- 16MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By