Problem-solving: 2017 6 provinces joint exam breaking up is a wish

Question Surface

First, consider what is the optimal strategy. In fact, this optimal strategy is one by one from big to small. Because a small one won't and won't affect the big one, it is obviously the best and necessary, so let's enumerate the factor and store it in vector, then we can directly simulate it to get $ 50pts $ (as if I had made water from Grandpa bi's data (?) Resulting in 80 PTS (FOG)

Consider the random press process before reaching $ K $ and set $ exp [I] $ (do not write this in the code, exp is a function in cmath) it also needs to meet the expectation for $ I $ times.

$ Exp [I] =\frac {I} {n} * 1 + \ frac {n-I} {n} * (exp [I] + exp [I + 1] + 1) $

$ \ Frac {I} {n} * 1 $ indicates that this time is correct and it is transferred directly, $ \ frac {n-I} {n} * (exp [I] + exp [I + 1] + 1) $ indicates that the error was reported this time, returns $ exp [I + 1] $ and transfers back $ exp [I] $.

Then, a recursive formula is obtained for simplifying and moving items.

$ Exp [I] = \ frac {(n-I) * (exp [I + 1] + 1)} {I} + 1 $

Remember to multiply $ n! $

 1 #include<cstdio> 2 #include<vector> 3 #include<cstring> 4 #include<algorithm> 5 using namespace std; 6 const int N=100005; 7 const long long mod=100003; 8 long long inv[N],lit[N],ex[N]; 9 long long n,k,cnt,ans,facs;10 vector<long long> vc[N];11 void prework()12 {13     facs=inv[0]=inv[1]=1;14     for(int i=2;i<=n;i++)15     {16         facs=facs*i%mod;17         inv[i]=((mod-mod/i)*inv[mod%i])%mod;18     }19 }20 int main()21 {22     scanf("%lld%lld",&n,&k),prework();23     for(int i=1;i<=n;i++)24         scanf("%lld",&lit[i]);25     for(int i=1;i<=n;i++)26         for(int j=i;j<=n;j+=i)27             vc[j].push_back(i);28     for(int i=n;i;i--)29         if(lit[i])30         {31             cnt++;32             for(int j=0;j<(int)vc[i].size();j++)33                 lit[vc[i][j]]^=1;34         }35     if(cnt<=k) 36         printf("%lld",cnt*facs%mod);37     else38     {39         ex[n]=1;40         for(int i=n-1;i>k;i--)41             ex[i]=((n-i)*(ex[i+1]+1)%mod*inv[i]%mod+1)%mod;42         for(int i=1;i<=cnt;i++)43             ans+=ex[i],ans%=mod;44         printf("%lld",(ans+k)*facs%mod);45     }46     return 0;47 }

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Problem-solving: 2017 6 provinces joint exam breaking up is a wish