线段树模板 poj3468

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 115916 Accepted: 36013Case 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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

模板：

#include<bits/stdc++.h>using namespace std;typedef long long LL;const int N =200011;LL tree[N<<2];LL lazy[N<<2];LL a[N];void updata(int n){    tree[n]=tree[n<<1]+tree[n<<1|1];}void pushdown(int l,int r,int n){    if(lazy[n])    {        int m=(l+r)>>1;        lazy[n<<1]=lazy[n<<1]+lazy[n];        lazy[n<<1|1]=lazy[n<<1|1]+lazy[n];        tree[n<<1]=tree[n<<1]+(m-l+1)*lazy[n];        tree[n<<1|1]=tree[n<<1|1]+(r-m)*lazy[n];        lazy[n]=0;    }}void build(int l,int r,int n){    lazy[n]=0;    if(l==r)    {        tree[n]=a[l];        return ;    }    int m=(l+r)>>1;    build(l,m,n<<1);    build(m+1,r,n<<1|1);    updata(n);    return ;}LL query(int l,int r,int n,int a,int b){    if(l>=a&&r<=b)        return tree[n];    pushdown(l,r,n);    int m=(l+r)>>1;    LL ans=0;    if(a<=m)        ans+=query(l,m,n<<1,a,b);    if(b>m)        ans+=query(m+1,r,n<<1|1,a,b);    return ans;}void add(int l,int r,int n,int a,int b,LL k){    if(l>=a&&r<=b)    {        lazy[n]+=k;        tree[n]=tree[n]+(r-l+1)*k;        return ;    }    pushdown(l,r,n);    int m=(l+r)>>1;    if(a<=m)        add(l,m,n<<1,a,b,k);    if(b>m)        add(m+1,r,n<<1|1,a,b,k);    updata(n);    return ;}int main(){    int n,m,i;    while(scanf("%d",&n)!=EOF)    {        scanf("%d",&m);        for(i=1;i<=n;i++)            scanf("%lld",&a[i]);        build(1,n,1);        char q;        int l,r;        LL k;        while(m--)        {            scanf(" %c%d%d",&q,&l,&r);            if(q=='Q')                printf("%lld\n",query(1,n,1,l,r));            else if(q=='C')            {                scanf("%lld",&k);                add(1,n,1,l,r,k);            }        }    }    return 0;}