7 条题解

  • 2
    @ 2023-11-13 18:35:40
    #include <bits/stdc++.h>
    using namespace std;
    int main()
    {
        int n;
        cin >> n;
        int a[105][105];
        for (int i = 1; i <= n; i++)
        {
            for (int j = 1; j <= i; j++)
            {
                if (j == 1 || i == j)
                    a[i][j] = 1;
                else
                    a[i][j] = a[i - 1][j] + a[i - 1][j - 1];
                cout << a[i][j] << " ";
            }
            cout << endl;
        }
    }
    
    • 2
      @ 2023-11-10 22:30:18
      #include <iostream>
      int main(){
          int n,a[25][25]; std::cin>>n;
          for(int i=1;i<=n;i++){
              for(int j=1;j<=i;j++){
                  if(j==1||j==i)a[i][j]=1;
                  else a[i][j]=a[i-1][j-1]+a[i-1][j];}}
          for(int i=1;i<=n;i++){
              for(int j=1;j<=i;j++)
                  std::cout<<a[i][j]<<" ";
              puts("");}
          return 0;}
      
      • 1
        @ 2024-3-23 20:20:36
        #include<cstdio>
        int a[21][21];
        int main()
        {
        int n;
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        a[i][1]=a[i][i]=1;//赋初值
        for(int i=1;i<=n;i++)
        for(int j=2;j<i;j++)//因为a[i][1]、a[i][i]已经赋值过了,所以循环是2~n-1
        a[i][j]=a[i-1][j]+a[i-1][j-1];
        for(int i=1;i<=n;i++)
        {
        for(int j=1;j<=i;j++)
        printf("%d ",a[i][j]);
        printf("\n");
        }
        }
        
        
        • 0
          @ 2024-4-20 22:04:19
          #include<iostream>
          using namespace std;
          int a[35][35];
          int main()
          {
              int n;
              cin >> n;
              for(int i=1;i<=n;i++)
              {
                  for(int j=1;j<=i;j++)
                  {
                      if(j==1 || j==i)
                      {
                          a[i][j]=1;
                      }
                      else
                      {
                          a[i][j]=a[i-1][j-1]+a[i-1][j];
                      }
                  }
              }
              for(int i=1;i<=n;i++)
              {
                  for(int j=1;j<=i;j++)
                  {
                      cout << a[i][j] << " ";
                  }
                  cout << endl;
              }
              return 0;
          }
          
          • 0
            @ 2024-3-22 19:59:09

            image

            部分代码:

            for(ll i=0;i<n;i++){
                f[i][0]=1;
                for(ll j=1;j<=i;j++)
                    f[i][j]=f[i-1][j]+f[i-1][j-1];
                for(ll j=0;j<=i;j++)
                    printf("%lld%c",f[i][j],j==i?'\n':' ');
            }
            

            upd. 将原代码替换为了老师课上讲的代码,虽然原代码好像没什么问题,但是套在这一题上好像就有问题了。还是背这个板子把

            • 0
              @ 2023-12-9 20:44:04
              #include<iostream>
              int n;
              int f[21][21];
              int main()
              {
                  std::cin>>n;
                  for(int i=1;i<=n;i+=1)for(int j=1;j<=i;j+=1)if(j==1||j==i)f[i][j]=1;else f[i][j]=f[i-1][j-1]+f[i-1][j];
                  for(int i=1;i<=n;i+=1)
                  {
                      for(int j=1;j<=i;j+=1)std::cout<<f[i][j]<<" ";
                      std::cout<<"\n";
                  }
              }
              
              • 0
                @ 2023-11-20 21:27:26

                超级简单易懂

                #include <bits/stdc++.h>
                using namespace std;
                int a[25][25];
                int main()
                {
                    int n;
                    cin >> n;
                    for(int i = 1;i <= n;i++)
                    {
                        for(int j = 1;j <= i;j++)
                        {
                            if(j == 1 ||j == i)
                            {
                                a[i][j] = 1;
                            }
                            else
                            {
                                a[i][j] = a[i-1][j-1]+a[i-1][j];
                            }
                            cout << a[i][j] << " ";
                        }
                        cout << endl;
                    }
                }
                
                • 1

                信息

                ID
                566
                时间
                1000ms
                内存
                256MiB
                难度
                4
                标签
                递交数
                353
                已通过
                152
                上传者