poj 3468 A Simple Problem with Integers（线段树——区间更新）

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

ps：裸的区间更新模板题，lazy思想（延迟更新）真的是lazy啊，如果不用那一段区间时就不会更新它的值，什么时候用什么时候才会更新

代码：

#include<stdio.h>

#define maxn 100010
typedef __int64 LL;
struct node
{
LL val,le,ri;
LL mid()
{
return (le+ri)>>1;
}
} tree[maxn*4];
LL a[maxn],add[maxn*4];

void Build(LL rt,LL left,LL right)
{
tree[rt].le=left;
tree[rt].ri=right;
if(left==right)
{
tree[rt].val=a[left];
return ;
}
LL mid=tree[rt].mid();
Build(rt<<1,left,mid);
Build(rt<<1|1,mid+1,right);
tree[rt].val=tree[rt<<1].val+tree[rt<<1|1].val;
}

void Pushdown(LL rt,LL m)
{
if(add[rt])
{
add[rt<<1]+=add[rt];
add[rt<<1|1]+=add[rt];
tree[rt<<1].val+=add[rt]*(m-(m>>1));//注意这里,想一下
tree[rt<<1|1].val+=add[rt]*(m>>1);//还有这里
add[rt]=0;
}
}

void Updata(LL rt,LL left,LL right,LL num)
{
if(tree[rt].le==left&&right==tree[rt].ri)
{
add[rt]+=num;
tree[rt].val+=num*(tree[rt].ri-tree[rt].le+1);
return ;
}
Pushdown(rt,tree[rt].ri-tree[rt].le+1);
LL mid=tree[rt].mid();
if(right<=mid)
Updata(rt<<1,left,right,num);
else if(left>mid)
Updata(rt<<1|1,left,right,num);
else
{
Updata(rt<<1,left,mid,num);
Updata(rt<<1|1,mid+1,right,num);
}
tree[rt].val=tree[rt<<1].val+tree[rt<<1|1].val;
}

LL Query(LL rt,LL left,LL right)
{
if(tree[rt].le==left&&tree[rt].ri==right)
return tree[rt].val;
Pushdown(rt,tree[rt].ri-tree[rt].le+1);
LL mid=tree[rt].mid();
if(right<=mid)
return Query(rt<<1,left,right);
else if(left>mid)
return Query(rt<<1|1,left,right);
else
return Query(rt<<1,left,mid)+Query(rt<<1|1,mid+1,right);
}

int main()
{
LL n,q,x,y,z;
char s[5];
scanf("%I64d%I64d",&n,&q);
for(LL i=1; i<=n; ++i)
scanf("%I64d",&a[i]);
Build(1,1,n);
for(LL i=1; i<=q; ++i)
{
scanf("%s",s);
if(s[0]=='Q')
{
scanf("%I64d%I64d",&x,&y);
printf("%I64d\n",Query(1,x,y));
}
else
{
scanf("%I64d%I64d%I64d",&x,&y,&z);
Updata(1,x,y,z);
}
}
return 0;
}