D. The Child and Sequence
Time Limit: 20 Sec
Memory Limit: 256 MB
题目连接
http://codeforces.com/contest/438/problem/D
Description
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.
Sample Input
5 5
1 2 3 4 5
2 3 5 4
3 3 5
1 2 5
2 1 3 3
1 1 3
Sample Output
8
5
HINT
题意
给你n个数,三个操作
1.输出[l,r]的和
2.将[l,r]中的数,对v取摸
3.把a[x]变成v
题解:
线段树
区间定值和区间和很简单
区间取摸的话,需要维护一个区间最大值,如果这个区间的最大值小于要取摸的数,那么就直接break就好了
代码
#include<iostream> #include<stdio.h> using namespace std; #define maxn 100005 struct node { int l,r; long long mx,sum; }a[maxn*4]; int d[maxn]; void build(int x,int l,int r) { a[x].l = l,a[x].r = r; if(l==r) { a[x].mx = a[x].sum = d[l]; return; } int mid = (l+r)/2; build(x<<1,l,mid); build(x<<1|1,mid+1,r); a[x].mx = max(a[x<<1].mx , a[x<<1|1].mx); a[x].sum = a[x<<1|1].sum + a[x<<1].sum; } void change(int x,int pos,long long val) { int l = a[x].l,r = a[x].r; if(l==r) { a[x].mx = a[x].sum = val; return; } int mid = (l+r)/2; if(pos<=mid) change(x<<1,pos,val); else change(x<<1|1,pos,val); a[x].mx = max(a[x<<1].mx,a[x<<1|1].mx); a[x].sum = a[x<<1].sum + a[x<<1|1].sum; } void mod(int x,int l,int r,long long val) { int L = a[x].l,R = a[x].r; if(a[x].mx<val)return; if(L==R) { a[x].sum%=val; a[x].mx%=val; return; } int mid = (L+R)/2; if(r<=mid) mod(x<<1,l,r,val); else if(l>mid) mod(x<<1|1,l,r,val); else mod(x<<1,l,mid,val),mod(x<<1|1,mid+1,r,val); a[x].sum = a[x<<1].sum + a[x<<1|1].sum; a[x].mx = max(a[x<<1].mx,a[x<<1|1].mx); } long long get(int x,int l,int r) { int L = a[x].l,R = a[x].r; if(L>=l&&R<=r) { return a[x].sum; } int mid = (L+R)/2; long long sum1 = 0,sum2 = 0; if(r<=mid) sum1 = get(x<<1,l,r); else if(l>mid) sum2 = get(x<<1|1,l,r); else sum1 = get(x<<1,l,mid),sum2 = get(x<<1|1,mid+1,r); return sum1 + sum2; } int main() { int n,q; scanf("%d%d",&n,&q); for(int i=1;i<=n;i++) scanf("%d",&d[i]); build(1,1,n); while(q--) { int op; scanf("%d",&op); if(op==1) { int x,y;scanf("%d%d",&x,&y); printf("%lld ",get(1,x,y)); } if(op==2) { int x,y,z;scanf("%d%d%d",&x,&y,&z); mod(1,x,y,z); } if(op==3) { int x,y;scanf("%d%d",&x,&y); change(1,x,y); } } }