Codevs-4919 Segment Tree Exercise 4 (interval plus a value and find the number of intervals divisible by K, segment tree + array maintenance)

Give you the number of n, there are two operations

1: Add x to all numbers within the interval [A, a]

2: Query interval [A, b] can be divisible by 7 of the number

Enter a description Input Description

The first line is a positive integer n, the next n rows n integers, and then a positive integer q, which represents the number of operations. The next Q line is a number of integers per line. If the first number is add, followed by 3 positive integer a,b,x, indicates that each number in the interval [a, a] increases by X, if it is count, indicating that the statistic interval [a, b] can be divisible by 7

Output description Output Description

One answer for each query output line

Sample input Sample Input

3 2 3 46count 1 3count 1 2add 1 3 2count 1 3add 1 3 3count 1 3

Sample output Sample Output

0

0

0

1

Data range and Tips Data Size & Hint

10%:1<n<=10,1<q<=10

30%:1<n<=10000,1<q<=10000

100%:1<n<=100000,1<q<=100000

The puzzle: and the ordinary line segment of the tree interval modification and interval summation is similar, but we use an array here to store each interval on the number of k after the sum of the number of each, the other operations and ordinary line tree is the same;

Reference code:

1#include <iostream>2#include <cstdio>3#include <cstring>4#include <cmath>5#include <cstdlib>6#include <algorithm>7#include <queue>8#include <stack>9#include <Set>Ten#include <vector> One#include <map> A using namespacestd; -typedefLong LongLL; - Const intinf=0x3f3f3f3f; the ConstLL inf=0x3f3f3f3f3f3f3f3fll; - Const intmaxn=1e5+Ten; - intn,q,c[maxn],a,b,x,temp[7]; - stringstr; +  - structnode{ +     intl,r,tag,mod[7]; A} tree[maxn<<2]; at  - voidBuildintPosintLintR) - { -tree[pos].l=l,tree[pos].r=r,tree[pos].tag=0; -      for(intI=0;i<7; i++) tree[pos].mod[i]=0; -     if(l==R) in     { -tree[pos].mod[c[l]%7]=1; to         return ; +     } -     intMid= (TREE[POS].L+TREE[POS].R) >>1; theBuild (pos<<1, l,mid); *Build (pos<<1|1, mid+1, R); $      for(intI=0;i<7; i++) tree[pos].mod[i]=tree[pos<<1].mod[i]+tree[pos<<1|1].mod[i];Panax Notoginseng } -  the voidChangeintPosintx) + { A      for(intI=0;i<7; i++) temp[(i+x)%7]=Tree[pos].mod[i]; the      for(intI=0;i<7; i++) tree[pos].mod[i]=Temp[i]; + } -  $ voidPushup (intPOS) $ { -      for(intI=0;i<7; i++)  -tree[pos].mod[i]=tree[pos<<1].mod[i]+tree[pos<<1|1].mod[i]; the } - Wuyi voidPushdown (intPOS) the { -     if(TREE[POS].L==TREE[POS].R)return ; Wutree[pos<<1].tag+=Tree[pos].tag; -tree[pos<<1|1].tag+=Tree[pos].tag; AboutChange (pos<<1, Tree[pos].tag), Change (pos<<1|1, Tree[pos].tag); $tree[pos].tag=0; - } -  - voidUpdateintPosintLintRintx) A { +     if(Tree[pos].tag) pushdown (POS); the     if(tree[pos].l==l&&tree[pos].r==R) -     { $tree[pos].tag+=X;change (pos,x); the         return ; the     } the     intMid= (TREE[POS].L+TREE[POS].R) >>1; the     if(r<=mid) Update (pos<<1, l,r,x); -     Else if(l>=mid+1) Update (pos<<1|1, l,r,x); in     ElseUpdate (pos<<1, l,mid,x), update (pos<<1|1, mid+1, r,x); the pushup (POS); the } About  the intQueryintPosintLintR) the { the     if(Tree[pos].tag) pushdown (POS); +     if(TREE[POS].L==L&AMP;&AMP;TREE[POS].R==R)returntree[pos].mod[0]; -     intMid= (TREE[POS].L+TREE[POS].R) >>1; the     if(R<=mid)returnQuery (pos<<1, l,r);Bayi     Else if(l>=mid+1)returnQuery (pos<<1|1, l,r); the     Else returnQuery (pos<<1, L,mid) +query (pos<<1|1, mid+1, R);  the } -  - intMain () the { theIos::sync_with_stdio (false); theCin>>N; the      for(intI=1; i<=n;i++) cin>>C[i]; -Build1,1, n); theCin>>Q; the      while(q--) the     {94Cin>>str; the         if(str[0]=='a') the         { theCin>>a>>b>>x;98Update1, a,b,x); About         } -         Else if(str[0]=='C')101         {102Cin>>a>>b;103Cout<<query (1, b) <<Endl;104         } the     }106     107     return 0;108}

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Codevs-4919 Segment Tree Exercise 4 (interval plus a value and find the number of intervals divisible by K, segment tree + array maintenance)