CF C. Inna and Candy Boxes

Test instructions: give you a string of length n containing only 1 and 0, W inquiry, each query input l,r, in [L,r] in the l+k-1, l+2*k-1 、...... R position must be 1, if not 1, into 1, recorded as an operation, the other place must be 0, Not 0 of the place to become 1, also recorded as an operation, and finally asked in the interval [l,r] at least several operations.

Idea: You can use a tree-like array to record how many 1 to the right of a place, and then preprocess out from 1, 2,. K for the beginning of the position of the i+c*k-1 in front of how many 1, the final answer is to take away the 1+ added 1, take away the number of 1 is equal to the number of existing 1-position the correct number of 1. The number of 1 added is equal to the total number of items to be added-1 of the correct number of positions.

1#include <cstdio>2#include <cstring>3#include <algorithm>4 #defineMAXN 1000105 using namespacestd;6 7 intn,k,w;8 CharSTR[MAXN];9 intA[MAXN];Ten intC[MAXN]; One intDP[MAXN]; A intSUM[MAXN]; -  - intLowbit (intx) the { -     returnx&-x; - } -  + voidInsertintXintd) - { +      while(x<MAXN) A     { atc[x]+=D; -x+=lowbit (x); -     } - } -  - intGetsum (intx) in { -     intans=0; to      while(x>0) +     { -ans+=C[x]; thex-=lowbit (x); *     } $     returnans;Panax Notoginseng } -  the  + intMain () A { the      while(SCANF ("%d%d%d", &n,&k,&w)! =EOF) +     { -scanf"%s", str+1); $memset (SUM,0,sizeof(sum)); $Memset (A,0,sizeof(a)); -Memset (c,0,sizeof(c)); -           for(intI=1; i<=n; i++) the          { -             if(str[i]=='1')Wuyi             { theInsert (I,1); -             } Wu          } -           for(intI=1; i<=k; i++) About          { $              intCnt=0; -               for(intC=1; i+c*k-1<=n; C++) -              { -cnt+= (str[i+c*k-1]-'0'); Adp[i+c*k-1]=CNT; +              } the          } -           for(intI=1; i<=w; i++) $          { the              intL,r; the              intcn1=0; thescanf"%d%d",&l,&R); the              intNum= (r-l+1)/K; -Ans+=getsum (R)-getsum (l1); inans-=dp[r]-dp[l-1]; theans+=num; theans-=dp[r]-dp[l-1]; Aboutprintf"%d\n", ans); the          } the     } the     return 0; +}

View Code

CF C. Inna and Candy Boxes