1 solutions
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0
C :
#include<stdio.h> 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 < i - 1;j++){ printf(" "); } //控住输出的* for(z=1;z <= n;z++){ printf("*"); } //负责换行 printf("\n"); } }
C++ :
#include <iostream> using namespace std; int main(){ int i,n,j; cin>>n; for(i=1;i<=n;i++){ for(j=1;j<=i-1;j++){ cout<<" "; } for(j=1;j<=n;j++){ cout<<"*"; } cout<<endl; } }
Python :
n= int(input()) for i in range (1, n + 1): for k in range (1,i): print(' ', end='') for j in range (1, n + 1): print('*', end = '') print();
- 1
Information
- ID
- 10877
- Time
- 1000ms
- Memory
- 16MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By