POJ 3468 A Simple Problem with Integers(线段树模板之区间增减更新 区间求和查询)
Time Limit: 5000MS | Memory Limit: 131072K | |
Total Submissions: 140120 | Accepted: 43425 | |
Case Time Limit: 2000MS |
Description
You have N integers, A1, A2, ... , AN. You need to deal with two kinds of operations. One type of operation is to add some given number to each number in a given interval. The other is to ask for the sum of numbers in a given interval.
Input
The first line contains two numbers N and Q. 1 ≤ N,Q ≤ 100000.
The second line contains N numbers, the initial values of A1, A2, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
Output
You need to answer all Q commands in order. One answer in a line.
Sample Input
10 5
1 2 3 4 5 6 7 8 9 10
Q 4 4
Q 1 10
Q 2 4
C 3 6 3
Q 2 4
Sample Output
4
55
9
15
Hint
#include<stdio.h>
#include<iostream>
#include<vector>
#include <cstring>
#include <stack>
#include <cstdio>
#include <cmath>
#include <queue>
#include <algorithm>
#include <vector>
#include <set>
#include <map>
#include<string>
#include<math.h>
#define max_v 100005
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
typedef long long LL;
LL sum[max_v<<],add[max_v<<];
struct node
{
int l,r;
int mid()
{
return (l+r)/;
}
}tree[max_v<<]; void push_up(int rt)//向上更新
{
sum[rt]=sum[rt<<]+sum[rt<<|];
}
void push_down(int rt,int m)//向下更新
{
if(add[rt])//若有标记,则将标记向下移动一层
{
add[rt<<]+=add[rt];
add[rt<<|]+=add[rt]; sum[rt<<]+=add[rt]*(m-(m>>));
sum[rt<<|]+=add[rt]*(m>>);
add[rt]=;//取消本层标记
}
}
void build(int l,int r,int rt)
{
tree[rt].l=l;
tree[rt].r=r;
add[rt]=; if(l==r)
{
scanf("%I64d",&sum[rt]);
return ;
} int m=tree[rt].mid();
build(lson);
build(rson);
push_up(rt);//向上更新
}
void update(int c,int l,int r,int rt)
{
if(tree[rt].l==l&&tree[rt].r==r)
{
add[rt]+=c;
sum[rt]+=(LL)c*(r-l+);
return ;
} if(tree[rt].l==tree[rt].r)
return ; push_down(rt,tree[rt].r-tree[rt].l+);//向下更新 int m=tree[rt].mid();
if(r<=m)
update(c,l,r,rt<<);
else if(l>m)
update(c,l,r,rt<<|);
else
{
update(c,l,m,rt<<);
update(c,m+,r,rt<<|);
}
push_up(rt);//向上更新
}
LL getsum(int l,int r,int rt)
{
if(tree[rt].l==l&&tree[rt].r==r)
return sum[rt]; push_down(rt,tree[rt].r-tree[rt].l+);//向下更新 int m=tree[rt].mid();
LL res=;
if(r<=m)
res+=getsum(l,r,rt<<);
else if(l>m)
res+=getsum(l,r,rt<<|);
else
{
res+=getsum(l,m,rt<<);
res+=getsum(m+,r,rt<<|);
}
return res;
}
int main()
{
int n,m;
while(~scanf("%d %d",&n,&m))
{
build(,n,);//1到n建树 rt为1 while(m--)
{
char str[];
int a,b,c;
scanf("%s",str);
if(str[]=='Q')
{
scanf("%d %d",&a,&b);
printf("%I64d\n",getsum(a,b,));
}else
{
scanf("%d %d %d",&a,&b,&c);
update(c,a,b,);
}
}
}
return ;
}
/*
区间更新:增减更新
区间查询:求和
*/
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