poj 3468 A Simple Problem with Integers 线段树,区间更新.

A Simple Problem with Integers,题目链接 ,click here

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

The sums may exceed the range of 32-bit integers.

Source

POJ Monthly–2007.11.25, Yang Yi

一道很经典的线段树区间更新，上浮还有下浮，套模板就能过的．

#include<stdio.h>
#include<algorithm>
#include<iostream>
using namespace std;
#define lson l,m,rt<<1       //左儿子;
#define rson m+1,r,rt<<1|1   //右儿子;
#define root 1,n,1           //建树的时候简便用的;
#define ll long long         //long long ;
const int N    =  100005;
ll sum[N<<2];   //建树数组;
ll add[N<<2];   //延迟标记,更新;

//上浮;
void pushup(int rt)
{
sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
//下沉；
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;
}
}

//建树;
void build(int l,int r,int rt)
{
add[rt]=0;
if(l==r)
{
scanf("%I64d",&sum[rt]);
return ;
}
int m=(l+r)>>1;
build(lson);
build(rson);
pushup(rt);//注意上浮;
}

//更新;
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);//注意下沉;
int m=(l+r)>>1;
//    if(L<=m) update(L,R,c,lson);
//    if(R>m)  update(L,R,c,rson);
//这里的判断用上边的还是用下边的都可以;
if(R<=m)
update(L,R,c,lson);
else if(L>m)
update(L,R,c,rson);
else
{
update( L ,m,c,lson);
update(m+1,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=(l+r)>>1;
ll ans=0;
if(L<=m) ans+=query(L,R,lson);
if(R>m)  ans+=query(L,R,rson);
return ans;
}
int main()
{
int n,q;
scanf("%d%d",&n,&q);
build(root);
char op[5];
while(q--)
{
int a,b,c;
scanf("%s",op);
if(op[0]=='Q')
{
scanf("%d%d",&a,&b);
printf("%I64d\n",query(a,b,root));
}
else
{
scanf("%d%d%d",&a,&b,&c);
update(a,b,c,root);
}
}
return 0;
}

很简单的一道题,硬是让我写了一下午,写了两遍,对自己的智商感到深深的忧桑~!!

一个位运算”<<”让我写成”>>”,改了好久…好坑!!一直RE.QAQ