HDU 5439 Aggregated counting

Main topic:

Add numbers to the array starting from 1

is to add this number by the number represented in the current location.

1 2 2 3 3 since the 4 is required, and the 4th position is 3, the following 3 4 are required

The problem is that given a n, find the last position n appears p, and then find the last position p appears to

Here you can consider finding g[i] to show how many times I have appeared

Think about it here, because G[i] is equivalent to the number of the I position appears

So g[i] also satisfies this sequence

F[i] Indicates the last position where I appears, i.e. the total number of 1~i

Back to calculate g[i] When you can consider is to find the first position in the f[] interval, if f[k-1]< i <= f[k]

So that means g[i] = k at this time

Then we can calculate the complexity of Logn g[n].

To calculate the final answer, consider that given N, find the last occurrence of P, median length p = g[1]+g[2]....+g[p]

Then find the corresponding ans, then each increase of g[i], will let the length of the whole sequence increase i*g[i]

I*g[i] can be understood so that the length of I is the number of g[i], so the total length is i*g[i]

So ans = Sigma (i*g[i]) i<=n

So for n <= 1e9

Then a rough calculation will find F[500000]>1e9

So g[n]<500000 as long as the violence to the top 500000 g[], f[]

So before the answer is calculated, find G[n] how much

g[n]= Lower_bound (f+1, f+n+1, N)-F

Then explain [1, g[n]-1] all the lengths in this section have been used

So before preprocessing this length prefix and sum[]

For each length I, the number of times he appears is f[i]-f[i-1]

Sigma (N*g[n]) f[i-1]<n<=f[i]-g[n] = i

Then the answer is i* arithmetic progression, remember to take the mold (⊙o⊙) OH

Then (G[n]-1, G[n]] this paragraph as long as the enumeration (G[N]-1, n] is ok

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <vector>5#include <queue>6 using namespacestd;7 #defineM 268 #defineN 5000009 #defineull unsigned long LongTen #definell Long Long One Const intMOD =1000000007; A intf[n+2], g[n+2], cnt[n+2]; -ll sum[n+2]; -  the intHash (intv) - { -     returnLower_bound (f+1, f+n+1, v)-F; - } +  - voidInit () + { Ag[1] =1, f[1] =1; atg[2] =2, f[2] =3; -      for(intI=3; I<=n; i++){ -G[i] = Lower_bound (f+1, F+i, I)-F; -F[i] = f[i-1]+G[i]; -     } - //for (int i=1; i<=100; i++) in //cout<<i<< "" <<g[i]<< "" <<f[i]<<endl; - //cout<<f[n]<<endl; to  +sum[1] =1; -      for(intI=2; I<=n; i++){ theSum[i] = sum[i-1]+ (LL) (f[i-1]+1+f[i]) * (f[i]-f[i-1])/2% MOD * (LL) I%MOD; *       //if (i<=10) cout<< "sum:" <<i<< "<<sum[i]<<endl; $     }Panax Notoginseng } - intMain () { the //freopen ("a.in", "R", stdin); + //freopen ("OUT.txt", "w", stdout); A  the init (); +     intT, N; -scanf"%d", &T); $      while(t--){ $scanf"%d", &n); -         intpos =Hash (n); -LL ret = sum[pos-1]; the          for(inti=f[pos-1]+1; I<=n; i++)//this interval is pos for each length -         {Wuyiret = (ret+ (ll) i*pos)%MOD; the         } -printf"%i64d\n", ret); Wu     } -     return 0; About}

HDU 5439 Aggregated counting