HDU 5312 Sequence (law problem)

Test instructions: The nth item of a sequence is 3*n* (n-1) +1, and N>=1, which now gives a positive integer m, asks at least how many of the numbers in the sequence are composed?

Idea: First, the 1th item of the sequence is 1, so any number can be constituted. But how much should it be at least? Deformation of the formula,6* (n (n-1)/2) +1, see the number of triangles N (n-1)/2, then it should be 6* (any natural number) +x=m is right, because up to 3 triangles can be composed of any natural number.

May I try m%6? Does this try to find x? Because any natural number is composed of up to 3, if it is k, then should be x>=k, do not forget there is a +1 item. X-k that part, it can only be done by the one.

There is a premise, m%6=0 how to do? It can't be made up of 0? That should be at least 1, so (m-1)%6 can be guaranteed not to be 0. Try M=13, then x= (13-1)%6+1=1, so can it really be composed of a number in a sequence? The order is 1,7,19 ... There seems to be no ... Tragedy! Should be 7+1+1+1+1+1+1 is, those 6 1 is to fill up. Similarly x=2 also need to ensure that there are two numbers in the sequence can be composed of M, otherwise x+=6 is the answer. And the 3 and above is not necessary, why I do not know ...

1#include <bits/stdc++.h>2 #defineLL Long Long3 #definePII pair<int,int>4 #defineINF 0x7f7f7f7f5 using namespacestd;6 Const intn=10100;7unordered_map<int,int>Mapp;8vector<int>seq;9 voidpre_cal ()Ten { One     intt=1; A      for(intI=1; i<20000&&t<=1000000000; i++) -     { -t=3*i* (I-1)+1; themapp[t]=1; - Seq.push_back (t); -     } - } +  - BOOLCalintm) + { A     intq=0, P=seq.size ()-1; at      while(q<=p) -     { -         intt=seq[q]+Seq[p]; -         if(t==m)return 1; -         if(t>m) p--; -         Elseq++; in     } -     return 0; to } +  - intMain () the { *     //freopen ("Input.txt", "R", stdin); $     intT, M;Panax Notoginseng pre_cal (); -Cin>>T; the      while(t--) +     { Ascanf"%d",&m); the         intk= (M-1)%6+1;//K is not more than 6. But the answer could be more than 6.  +  -         if(k>=3) printf ("%d\n", k); $         Else if(k==1)//Special Award $         { -             if(Mapp[m]) printf ("1\n"); -             Elseprintf"%d\n", k+=6); the         } -         Else if(k==2)//Special AwardWuyi         { the             if(Cal (M)) printf ("2\n"); -             Elseprintf"%d\n", k+=6); Wu         } -     } About     return 0; $}

AC Code

HDU 5312 Sequence (law problem)