1 solutions
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0
C :
#include<stdio.h> int main(){ int n,i,j,z; scanf("%d",&n); //负责行 for(i = 1;i <= n;i++){ //负责空格 for(j = 1;j <= n - i;j++){ printf(" "); } //负责打印数字 for(z=1;z <= 2 * i - 1;z++){ printf("%d",z); } //负责换行 printf("\n"); } return 0; }C++ :
#include<iostream> #include<cmath> using namespace std; int main() { char c; int i,t,n,j; cin>>n; for(i = 1;i <= n;i++){ for(j = 1;j <=n - i;j++){ cout<<' '; } for(j = 1;j <=2 * i - 1;j++){ cout<<j; } cout<<endl; } }Python :
n = int(input()) for i in range(1, n + 1): for j in range(1, n - i + 1): print(' ', end='') for j in range(1, 2 * i - 1 + 1): print(j, end='') print()
- 1
Information
- ID
- 1260
- Time
- 1000ms
- Memory
- 16MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By