4 条题解

  • 1
    @ 2026-2-14 19:45:51
    #include<bits/stdc++.h>
    #pragma GCC optimize(2)
    #define For(i,a,b) for(register ll i=a;i<=b;i++)
    #define Down(i,a,b) for(register ll i=a;i>=b;i--)
    using namespace std;
    typedef long long ll;
    typedef string str;
    typedef double db;
    const ll inf=0x3f3f3f3f3f3f3f3f;
    inline ll read(){
        ll x=0,f=1;
        char c=getchar();
        while(!isdigit(c)) f=(c=='-')?-1:f,c=getchar();
        while(isdigit(c)) x=x*10+c-'0',c=getchar();
        return x*f;
    }
    inline ll write(ll x){
    	if(x<0){
    		putchar('-');
    		x=-x;
    	}
    	if(x>9) write(x/10);
    	putchar(x%10+'0');
    }
    ll n,a[1000100];
    int main(){
    	cin>>n;
    	a[1]=1;
    	a[2]=2;
    	For(i,3,n) a[i]=(a[i-1]+a[i-2])%15746;
    	cout<<a[n];
        return 0;
    }
    

    猪CTJer HAPPY

    • 0
      @ 2025-11-8 18:03:58
      #include<bits/stdc++.h>
      #define ll long long
      #pragma GCC optimize(2)
      #define un unsigned
      #define int long long
      #define db double
      #define st string
      #define ct const
      #define xh(a,b,c) for(int a=b;a<=c;a++)
      #define wx while(1)
      #define dn(a,b,c) for(int a=b;a>=c;a--)
      using namespace std;
      int f[1000010],n;
      signed main(){
      	cin>>n;
      	f[1]=1,f[2]=2;
      	xh(i,3,n)f[i]=(f[i-1]+f[i-2])%15746;
      	cout<<f[n];
      	return 0;
      }
      
      • -1
        @ 2025-7-14 13:47:24

        #include<bits/stdc++.h> using namespace std; long long n,i,a[1050000]; int main(){ ios::sync_with_stdio(0); cin.tie(0); cout.tie(0); cin>>n; a[1]=1; a[2]=2; for(i=3;i<=n;i++) a[i]=(a[i-1]+a[i-2])%15746; cout<<a[n]; return 0; }

        • -2
          @ 2022-7-29 9:54:27

          就是斐波那契数列

          #include<bits/stdc++.h>
          using namespace std;
          long n,i,a[1000500];
          int main()
          {
             cin>>n;
             a[1]=1;
             a[2]=2;
             for(i=3;i<=n;i++)a[i]=(a[i-1]+a[i-2])%15746;
             cout<<a[n];
             return 0;
          }
          • 1

          信息

          ID
          55
          时间
          1000ms
          内存
          256MiB
          难度
          4
          标签
          (无)
          递交数
          71
          已通过
          34
          上传者