1 条题解

  • 0
    @ 2023-6-11 12:16:28

    C :

    #include<stdio.h>
    int a[100][100];
    int main(){
        int n,i,j;
        scanf("%d",&n);
        for(j=1;j<=n;j++){
            for(i=n;i>=1;i--){
                if(i==j){
                    a[i][j] = i;
                }else{
                    a[i][j] = a[i+1][j] + a[i][j-1];
                }
            }
        }
        for(i=1;i<=n;i++){
            for(j=1;j<=n;j++){
                if(i>j){
                    printf("%5s","");
                }else{
                    printf("%5d",a[i][j]);
                }
            }
            printf("\n");
        }
        return 0;
    }
    
    

    C++ :

    #include <bits/stdc++.h>
    using namespace std;
    
    int main(){
    	int a[20][20] = {0};
    	int i,j,n;
    	cin>>n;
    	
    	for (i = 0;i < n;i++){
    		a[i][i] = i + 1;
    	}
    	
    	for (i = 0;i < n - 1;i++){
    		for (j = 0;j < n - i - 1;j++){
    			a[j][j + i + 1] = a[j][j + i] + a[j + 1][j + i + 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)]
    
    # 先赋值左对角线
    for i in range(n):
        a[i][i] = i + 1
    # 以列为大方向,从1~n列
    for j in range(1, n):
        # 以行为小方向,从j-1~0
        for i in range(j - 1, -1, -1):
            a[i][j] = a[i + 1][j] + a[i][j - 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
    2141
    时间
    1000ms
    内存
    16MiB
    难度
    (无)
    标签
    递交数
    0
    已通过
    0
    上传者