CodeForces - 438D: The Child and Sequence(势能线段树)
At the children's day, the child came to Picks's house, and messed his house up. Picks was angry at him. A lot of important things were lost, in particular the favorite sequence of Picks.
Fortunately, Picks remembers how to repair the sequence. Initially he should create an integer array a[1], a[2], ..., a[n]. Then he should perform a sequence of m operations. An operation can be one of the following:
- Print operation l, r. Picks should write down the value of
.
- Modulo operation l, r, x. Picks should perform assignment a[i] = a[i] mod x for each i (l ≤ i ≤ r).
- Set operation k, x. Picks should set the value of a[k] to x (in other words perform an assignment a[k] = x).
Can you help Picks to perform the whole sequence of operations?
Input
The first line of input contains two integer: n, m (1 ≤ n, m ≤ 105). The second line contains n integers, separated by space: a[1], a[2], ..., a[n] (1 ≤ a[i] ≤ 109) — initial value of array elements.
Each of the next m lines begins with a number type .
- If type = 1, there will be two integers more in the line: l, r (1 ≤ l ≤ r ≤ n), which correspond the operation 1.
- If type = 2, there will be three integers more in the line: l, r, x (1 ≤ l ≤ r ≤ n; 1 ≤ x ≤ 109), which correspond the operation 2.
- If type = 3, there will be two integers more in the line: k, x (1 ≤ k ≤ n; 1 ≤ x ≤ 109), which correspond the operation 3.
Output
For each operation 1, please print a line containing the answer. Notice that the answer may exceed the 32-bit integer.
Examples
5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3
8
5
10 10
6 9 6 7 6 1 10 10 9 5
1 3 9
2 7 10 9
2 5 10 8
1 4 7
3 3 7
2 7 9 9
1 2 4
1 6 6
1 5 9
3 1 10
49
15
23
1
9
Note
Consider the first testcase:
- At first, a = {1, 2, 3, 4, 5}.
- After operation 1, a = {1, 2, 3, 0, 1}.
- After operation 2, a = {1, 2, 5, 0, 1}.
- At operation 3, 2 + 5 + 0 + 1 = 8.
- After operation 4, a = {1, 2, 2, 0, 1}.
- At operation 5, 1 + 2 + 2 = 5.
题意:给出数组,有三种操作,分别是区间求和,区间取模 ,单点修改。
思路:一个点被取模,那么其大小减半,所以一个数最多被操作log次,这样的话就不难想到势能线段树。
#include<bits/stdc++.h> #define ll long long using namespace std; const int maxn=100010; int Mx[maxn<<2]; ll sum[maxn<<2]; void build(int Now,int L,int R) { if(L==R){ scanf("%d",&Mx[Now]); sum[Now]=Mx[Now]; return ; } int Mid=(L+R)>>1; build(Now<<1,L,Mid); build(Now<<1|1,Mid+1,R); Mx[Now]=max(Mx[Now<<1],Mx[Now<<1|1]); sum[Now]=sum[Now<<1]+sum[Now<<1|1]; } ll query(int Now,int L,int R,int l,int r){ if(l<=L&&r>=R) return sum[Now]; int Mid=(L+R)>>1; ll res=0; if(l<=Mid) res+=query(Now<<1,L,Mid,l,r); if(r>Mid) res+=query(Now<<1|1,Mid+1,R,l,r); return res; } void change(int Now,int L,int R,int pos,int val) { if(L==R){ Mx[Now]=val; sum[Now]=val; return ; } int Mid=(L+R)>>1; if(pos<=Mid) change(Now<<1,L,Mid,pos,val); else change(Now<<1|1,Mid+1,R,pos,val); Mx[Now]=max(Mx[Now<<1],Mx[Now<<1|1]); sum[Now]=sum[Now<<1]+sum[Now<<1|1]; } void modp(int Now,int L,int R,int l,int r,int P) { if(Mx[Now]<P) return ; if(L==R) { Mx[Now]%=P; sum[Now]=Mx[Now]; return ; } int Mid=(L+R)>>1; if(l<=Mid) modp(Now<<1,L,Mid,l,r,P); if(r>Mid) modp(Now<<1|1,Mid+1,R,l,r,P); Mx[Now]=max(Mx[Now<<1],Mx[Now<<1|1]); sum[Now]=sum[Now<<1]+sum[Now<<1|1]; } int main() { int N,M,opt,L,R,P; scanf("%d%d",&N,&M); build(1,1,N); while(M--){ scanf("%d",&opt); if(opt==1) { scanf("%d%d",&L,&R); printf("%I64d\n",query(1,1,N,L,R)); } else if(opt==2){ scanf("%d%d%d",&L,&R,&P); modp(1,1,N,L,R,P); } else { scanf("%d%d",&L,&R); change(1,1,N,L,R); } } return 0; }