数据结构--线段树--lazy延迟操作

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 53749 Accepted: 16131Case 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

//更新某一段区域的时候，采用延迟标记~~
#include "cstdio" //poj 3468 lazy操作#include "cstring"#include "iostream"using namespace std;#define N 100005#define LL long longstruct node{    int x,y;    LL sum;    LL add;   //记录以当前节点为根节点的树中需要增加的值}a[3*N];void Build(int t,int x,int y){    a[t].x = x;     a[t].y = y;    a[t].sum = a[t].add = 0;    if(a[t].x == a[t].y)  //到了叶子节点    {        scanf("%lld",&a[t].sum);        return ;    }    int mid = (a[t].x + a[t].y)/2;    Build(t<<1,x,mid);    Build(t<<1|1,mid+1,y);    a[t].sum = a[t<<1].sum + a[t<<1|1].sum;}void Push_down(int t)  //将add(增值)向下推一级{    LL add = a[t].add;    a[t<<1].add += add;    a[t<<1|1].add += add;    a[t<<1].sum += add*(a[t<<1].y-a[t<<1].x+1);    a[t<<1|1].sum += add*(a[t<<1|1].y-a[t<<1|1].x+1);    a[t].add = 0;}LL Query(int t,int x,int y){    if(a[t].x==x &&a[t].y==y)        return a[t].sum;    Push_down(t);    int mid = (a[t].x + a[t].y)/2;    if(y<=mid)        return Query(t<<1,x,y);    if(x>mid)        return Query(t<<1|1,x,y);    else        return Query(t<<1,x,mid) + Query(t<<1|1,mid+1,y);}void Add(int t,int x,int y,int k){    if(a[t].x==x && a[t].y==y) //不在推到叶子节点，t下的子孙要增加的值存入a[t].add中    {        a[t].add += k;        a[t].sum += (a[t].y-a[t].x+1)*k;        return ;    }    a[t].sum += (y-x+1)*k;    Push_down(t);    int mid = (a[t].x+a[t].y)/2;    if(y<=mid)        Add(t<<1,x,y,k);    else if(x>mid)        Add(t<<1|1,x,y,k);    else    {        Add(t<<1,x,mid,k);        Add(t<<1|1,mid+1,y,k);    }}int main(){    int n,m;    char ch;    int x,y,k;    scanf("%d %d",&n,&m);    Build(1,1,n);    while(m--)    {        getchar();        scanf("%c %d %d",&ch,&x,&y);        if(ch=='Q')            printf("%lld\n",Query(1,x,y));        else        {            scanf("%d",&k);            Add(1,x,y,k);        }    }    return 0;}