POJ 3468 线段树 解题报告

A Simple Problem with Integers

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

【解题报告】

区间修改区间查询线段树。

写一下练练手

代码如下：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define maxn 100010
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1

long long add[maxn<<2],sum[maxn<<2];
int n,q;

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]+=(long long)c;
sum[rt]+=(long long)c*(r-l+1);
return;
}
pushdown(rt,r-l+1);
int m=(r+l)>>1;
if(L<=m) update(L,R,c,lson);
if(R>m) update(L,R,c,rson);
pushup(rt);
}
long long 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=(l+r)>>1;
long long ret=0;
if(L<=m) ret+=query(L,R,lson);
if(R>m) ret+=query(L,R,rson);
return ret;
}
int main()
{
scanf("%d%d",&n,&q);
build(1,n,1);
while(q--)
{
char opt[5];
int a,b,c;
scanf("%s",opt);
if(opt[0]=='Q')
{
scanf("%d%d",&a,&b);
printf("%lld\n",query(a,b,1,n,1));
}
if(opt[0]=='C')
{
scanf("%d%d%d",&a,&b,&c);
update(a,b,c,1,n,1);
}
}
return 0;
}