6 solutions

  • 3
    @ 2023-9-21 15:27:26

    #include<stdio.h>

    int main() {

    int n,a[10000007];

    scanf("%d",&n);

    a[1]=1,a[2]=1;

    if(n= =1 || n= =2){printf("1");}

    else {

    for(int i=3;i<=n;i++)

    {

    a[i]=a[i-1]%10007+a[i-2]%10007;

    }

    printf("%d",a[n]%10007);

    }

    return 0;

    }

    • 1
      @ 2022-4-4 17:15:53

      C++

      用递推写了一个,怕超时

      #include <iostream>
      using namespace std;
      int main()
      {
          int d1=1,d2=1,d3;
          int n = 0;
          cin >> n;
          int i = 2;
          if(n <= 2)  cout << d1 << endl;
          else
          {
              while(i < n)
              {
                  d3 = (d1+d2)%10007;
                  d1 = d2;
                  d2 = d3; 
                  i++;
              }
              cout << d3 << endl;
          }
          return 0;
      }
      
      • 1
        @ 2022-1-26 20:44:41
        #include<stdio.h>
        int main(){
        int f1=1,f2=1;
        int fn;
        int sum = 0;
        scanf("%d",&fn);
        if(fn<3){
        printf("1");
        }
        else{
        for(int i=3; i<=fn; i++){
        sum = (f1+f2)%10007;
        f1 = f2;
        f2 = sum;
        }
        printf("%d",sum);
        }
        return 0;
        }
        
        • 0
          @ 2023-12-18 17:26:20

          #include<stdio.h> int fb(int x) { int t,t1=1,t2=1; if(x1) return 1; else if(x2) return 1; else { for(int i=2;i<x;i++) { t=(t1+t2)%10007; t1=t2; t2=t; } return t; } } int main() { long long m; scanf("%d",&m); printf("%d",fb(m)); return 0; }

          • 0
            @ 2023-10-25 21:23:23
            #include <stdio.h>
            #pragma warning (disable:4996)
            
            int main() {
                int n;
                scanf("%d", &n);
                long long a = 0, b = 1, c;
                for (int i = 2; i <= n; i++) 
                {
                    c = (a + b) % 10007;
                    a = b;
                    b = c;
                }
                printf("%lld", b);
                    return 0;
            }
            
            
            • 0
              @ 2022-3-7 19:58:37
              #include <stdio.h>
              #include <algorithm>
              #include <math.h>
              #include <vector>
              using namespace std;
              vector<int> a;
              int main() {
                  int n,i=1;
                  a.push_back(1);
                  a.push_back(1);
                  scanf("%d",&n);
                  if(n==1||n==2)
                  {
                      printf("1\n");
                      return 0;
                  }
                  for(int i=2;i<n;i++)
                  {
                      int tmp=(a[i-1]%10007+a[i-2]%10007)%10007;
                      a.push_back(tmp);
                  }
                  printf("%d\n",a[n-1]);
                  return 0;
              }
              
              • 1

              Information

              ID
              38
              Time
              1000ms
              Memory
              256MiB
              Difficulty
              8
              Tags
              # Submissions
              1841
              Accepted
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              Uploaded By