hud-5475 an easy problem (segment tree)

http://acm.hdu.edu.cn/showproblem.php?pid=5475

Test instructions

At the beginning of the original number is 1 in order to give you the Q operation, and then there are two types of operations.

1 means that the number after the last operation is multiplied by the number given to you, then the remainder, then the output, 2 represents the addition of the operation, dividing the current number by the specified number of previous

Then modulo the M.

Ideas:

This problem is written with a line segment tree, the breakpoint is updated, and the complexity is n (log (n));

Code:

1#include <stdio.h>2#include <algorithm>3#include <iostream>4#include <string.h>5#include <stdlib.h>6typedefLong LongLL;7 voidUp (LL k,ll l);8 voidBuild (LL l,ll r,ll k);9 ll que (ll l,ll r,ll k,ll aa,ll dd);Tentypedefstructpp One { A LL x; - LL y; - LL ID; the LL t; - } SS; -typedefstructtree1 - { + LL x; - LL y; + LL ID; A LL cou; at } SD; - LL m,n; -SS cnt[100005]; - intflag[4*100005]; -SD tree[4*100005]; - using namespacestd; in intMainvoid) - { to LL n,i,j,k,p,q; +scanf"%lld",&n); -      for(i=1; i<=n; i++) the     { *scanf"%lld%lld",&n,&M); $memset (Tree,0,sizeof(tree));Panax Notoginseng          for(j=1; j<=n; J + +) -         { thescanf"%lld%lld",&cnt[j].x,&cnt[j].y); +Cnt[j].id=J; A             if(cnt[j].x==1) the             { +CNT[J].T=CNT[J].Y;//the number to multiply by 1 o'clock operation.  -             } $             Elsecnt[j].t=1;//operation 2 o'clock multiplied by the number equivalent to 1; $         } -Build1N0);//achievements (because each operation has a corresponding operation, so the operation 2 is also put together, the equivalent of the number to multiply is 1) -printf"Case #%lld:\n", i); the          for(j=1; j<=n; J + +) -         {Wuyi             if(cnt[j].x==1) the             { -LL Dd=que (1J0,1, N);//when 1 o'clock ask for some Wuprintf"%lld\n", DD); -             } About             Else if(cnt[j].x==2) $             { -Up (FLAG[CNT[J].Y],J);//when the 2 o'clock breakpoint is updated -LL Dd=que (1J0,1, N);//ask for some -printf"%lld\n", DD); A             } +         } the     } -     return 0; $ } the voidBuild (LL l,ll r,ll k)//Achievements the { thetree[k].x=l; thetree[k].y=R; -     if(l==R) in     { thetree[k].cou=cnt[l].t%M; theflag[l]=K; About         return; the     } the     Else the     { +Build (L, (l+r)/2,2*k+1); -Build ((L+R)/2+1R2*k+2); theTree[k].cou= (tree[2*k+1].cou*tree[2*k+2].cou)%M;Bayi     } the } the voidUp (LL k,ll L)//Breakpoint Update - { -tree[k].cou=1;//the point to be deleted is the equivalent of a multiply thek= (K-1)/2; the      while(k>=0)//update up to root node the     { theTree[k].cou= (tree[2*k+1].cou*tree[2*k+2].cou)%M; -         if(k==0) the         { the              Break; the         }94k= (K-1)/2; the     } the } thell que (ll l,ll r,ll k,ll aa,ll dd)//Ask98 { About     if(l>dd| | r<aa) -     {101         return 1;102     }103     Else if(l<=aa&&r>=dd)104     { the         returnTree[k].cou;106     }107     Else108     {109LL Nx=que (L,r,2*k+1, AA, (AA+DD)/2); theLL Ny=que (L,r,2*k+2, (AA+DD)/2+1, DD);111         return(nx*ny)%M; the     }113  the}

hud-5475 an easy problem (segment tree)