POJ 3468A Simple Problem with Integers(线段树)

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本文介绍了一种使用线段树解决区间更新与查询问题的方法。具体包括如何构建线段树、进行区间加法操作及求解区间和。通过样例输入输出展示了算法的有效性。

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 111222		Accepted: 34630
Case Time Limit: 2000MS

Description

You have N integers, A₁, A₂, ... , A_N. 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, ... , A_N. -1000000000 ≤ A_i ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of A_a, A_a₊₁, ... , A_b. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of A_a, A_a₊₁, ... , A_b.

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

想法：

Q a b表示询问a到b区间和

C a b c表示a到b区间都加c

线段树模板

代码：

#include<iostream>
#include<cstdio>
using namespace std;
typedef long long ll;
const int N=1e5+10;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
ll sum[N<<2];
ll add[N<<2];
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
void pushdown(int rt,int m)
{
if(add[rt])
{
add[rt<<1]+=add[rt];
add[rt<<1|1]+=add[rt];
sum[rt<<1]+=add[rt]*(m-(m>>1));
sum[rt<<1|1]+=add[rt]*(m>>1);
add[rt]=0;
}
}
void build(int l,int r,int rt)
{
add[rt]=0;
if(l==r)
{
scanf("%lld",&sum[rt]);
return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);
}
void update(int L,int R,int c,int l,int r,int rt)
{
if(L<=l&&R>=r)
{
add[rt]+=c;
sum[rt]+=(ll)c*(r-l+1);
return ;
}
pushdown(rt,r-l+1);
int m=(l+r)>>1;
if(L<=m) update(L,R,c,lson);
if(R>m) update(L,R,c,rson);
pushup(rt);
}
ll query(int L,int R,int l,int r,int rt)
{
if(L<=l&&R>=r)
return sum[rt];
pushdown(rt,r-l+1);
int m=(r+l)>>1;
ll res=0;
if(L<=m) res+=query(L,R,lson);
if(R>m) res+=query(L,R,rson);
return res;
}
int main()
{
int n,m;
while(~scanf("%d%d",&n,&m))
{
build(1,n,1);
while(m--)
{
char ch[2];
int a,b,c;
scanf("%s",ch);
if(ch[0]=='Q')
{
scanf("%d%d",&a,&b);
printf("%lld\n",query(a,b,1,n,1));
}
else
{
scanf("%d%d%d",&a,&b,&c);
update(a,b,c,1,n,1);
}
}
}
return 0;
}