Codeforces Round #740 (Div. 2, based on VK Cup 2021 - Final (Engine)) D1. Up the Strip (simplified v

• 题意：给你一个数\(x\),每次有两种操作可以选择，一是从\(x\)跳到\([1,x-1]\)的任意一个数，二是跳到\(\lfloor \frac{x}{z} \rfloor\ \ (z \in[2,x])\).问你从\(x\)到一有多少种方案.

• 题解：首先很容易写出dp公式，\(dp[t]=\sum_{i=t-1}^{t}dp[i]+\sum_{j=2}^{t}dp[\lfloor \frac{t}{j} \rfloor]\),前一个式子直接前缀和就能求出，后面的式子看到下取整求和，不难想到可以用数论分块\(O(\sqrt n)\)得出,这样就可以\(O(n \sqrt n)\)解决此题。

• 代码:

  #include <bits/stdc++.h>
  #define ll long long
  #define fi first
  #define se second
  #define pb push_back
  #define me memset
  #define rep(a,b,c) for(int a=b;a<=c;++a)
  #define per(a,b,c) for(int a=b;a>=c;--a)
  const int N = 1e6 + 10;
  const int mod = 1e9 + 7;
  const int INF = 0x3f3f3f3f;
  using namespace std;
  typedef pair<int,int> PII;
  typedef pair<ll,ll> PLL;
  ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
  ll lcm(ll a,ll b) {return a/gcd(a,b)*b;}
  
  ll n,m;
  ll dp[N];
  
  int main() {
      scanf("%lld %lld",&n,&m);
      dp[n]=1;
      ll sum=0;
      for(int i=n;i>=1;--i){
          dp[i]=(dp[i]+sum)%m;
          for(int l=2,r=0;l<=i;l=r+1){
              r=min(i,i/(i/l));
              dp[i/l]+=dp[i]*(r-l+1);
              dp[i/l]%=m;
          }
          sum=(sum+dp[i])%m;
      }
      printf("%lld\n",dp[1]);
      return 0;
  }
  
  

标签：740,Engine,const,int,ll,Up,sum,dp,define
来源： https://www.cnblogs.com/lr599909928/p/15206142.html