POJ 3468 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

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

一道线段树的模板题，题目大意：建立一棵线段树，支持区间加法，区间求和。

#include<iostream>#include<cstdio>using namespace std;const int maxn=100000;struct node{    int l,r;    long long sum,tmp;}tree[4*maxn+1];int n,q,data[maxn+1];void build(int l,int r,int num){    tree[num].l=l;    tree[num].r=r;    if(l==r)    {        tree[num].sum=data[l];        return ;    }    int mid=(l+r)/2;    build(l,mid,2*num);    build(mid+1,r,2*num+1);    tree[num].sum=tree[2*num].sum+tree[2*num+1].sum;}void update(int num){    tree[2*num].tmp+=tree[num].tmp;    tree[2*num].sum+=tree[num].tmp*(tree[2*num].r-tree[2*num].l+1);    tree[2*num+1].tmp+=tree[num].tmp;    tree[2*num+1].sum+=tree[num].tmp*(tree[2*num+1].r-tree[2*num+1].l+1);    tree[num].tmp=0;}void add(int l,int r,int x,int num){    if(tree[num].l>r||tree[num].r<l)        return ;    if(tree[num].l>=l&&tree[num].r<=r)    {        tree[num].tmp+=x;        tree[num].sum+=x*(tree[num].r-tree[num].l+1);        return ;    }    if(tree[num].tmp)        update(num);    add(l,r,x,2*num);    add(l,r,x,2*num+1);    tree[num].sum=tree[2*num].sum+tree[2*num+1].sum;}long long find_sum(int l,int r,int num){    if(tree[num].l>r||tree[num].r<l)        return 0;    if(tree[num].l>=l&&tree[num].r<=r)        return tree[num].sum;    if(tree[num].tmp)        update(num);    return find_sum(l,r,2*num)+find_sum(l,r,2*num+1);}int main(){    char ch;    scanf("%d%d",&n,&q);    for(int i=1;i<=n;++i)        scanf("%d",data+i);    build(1,n,1);    for(int i=1;i<=q;i++)    {        int a,b,c;        cin>>ch;        if(ch=='C')        {            scanf("%d%d%d",&a,&b,&c);            add(a,b,c,1);        }        else        {            scanf("%d%d",&a,&b);            printf("%lld\n",find_sum(a,b,1));        }    }    return 0;}