1 条题解

  • 0
    @ 2023-6-11 12:16:28

    C :

    #include<stdio.h> 
    int main(){
    	int n,i,j,k;
    	scanf("%d",&n);
    	int a[n][n];
    	for(i=0;i<n;i++){
    		k=1;
    		for(j=0;j<n-i-1;j++){
    			a[i][j]=k;
    		}
    		for(j=n-i-1;j<n;j++){
    			a[i][j]=k++;
    		}
    	}
    		for(i=0;i<n;i++){
    			for(j=0;j<n;j++){
    				if(i+j>=n-1){
    					printf("%5d",a[i][j]);
    				}else{
    					printf("%5s","");
    				}
    			}
    		printf("\n");
    	}
    	return 0;
    }
    

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    int main(){
    	int a[100][100] = {0};
    	int i,j,n;
    	cin>>n;
    	
    	for (i = 0;i < n;i++){
    		for (j = 0;j <= i;j++){
    			a[i][j + n - i - 1] = j + 1;
    		}
    	}
    	
    	for (i = 0;i < n;i++){
    		for (j = 0;j < n;j++){
    			if (a[i][j] == 0){
    				cout<<setw(5)<<" ";
    			}else{
    				cout<<setw(5)<<a[i][j];
    			}
    		}
    		cout<<endl;
    	}
    }
    

    Python :

    n = int(input())
    a = [[0 for j in range(n)] for i in range(n)]
    
    for i in range(n):
        a[i][n-i-1] = 1
        for j in range(n-i, n):
            a[i][j] = a[i][j - 1] + 1
    
    for i in range(n):
        for j in range(n):
            if a[i][j]==0:
                print('%5s' % '', end='')
            else:
                print('%5d' % a[i][j], end='')
        print()
    
    • 1

    信息

    ID
    2142
    时间
    1000ms
    内存
    16MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者