A Simple Problem with Integers（线段树）

题目：
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

题解：线段树板子题，区间查询区间修改。

代码：

#include<stdio.h>
#include<stdlib.h>
#include<math.h>
#include<string.h>
#include<algorithm>
#include<iostream>
#include<queue>
#include<stack>
#include<cstdio>
#include<cstring>
#include<vector>
#include<cmath>
#include<map>
#define exp 1e-9
#define PI acos(-1.0)
#define INF 0x3f3f3f3f
using namespace std;
typedef unsigned long long LL;
int n,p,a,b,m,x,y,xx,yy;
LL ans;
struct node
{
    LL l,r,w,f;
}tree[800001];
void build(int k,int ll,int rr)//建树
{
    tree[k].l=ll,tree[k].r=rr;
    if(tree[k].l==tree[k].r)
    {
        scanf("%lld",&tree[k].w);
        return;
    }
    int m=(ll+rr)/2;
    build(k*2,ll,m);
    build(k*2+1,m+1,rr);
    tree[k].w=tree[k*2].w+tree[k*2+1].w;
}
void add(int k)//单点修改
{
    if(tree[k].l==tree[k].r)
    {
        tree[k].w+=y;
        return;
    }
    int m=(tree[k].l+tree[k].r)/2;
    if(x<=m) add(k*2);
    else add(k*2+1);
    tree[k].w=tree[k*2].w+tree[k*2+1].w;
}
void qu(int k)//单点修改
{
    if(tree[k].l==tree[k].r)
    {
        tree[k].w-=yy;
        return;
    }
    int m=(tree[k].l+tree[k].r)/2;
    if(xx<=m) qu(k*2);
    else qu(k*2+1);
    tree[k].w=tree[k*2].w+tree[k*2+1].w;
}
void down(int k)//标记下穿
{
    tree[k*2].f+=tree[k].f;
    tree[k*2+1].f+=tree[k].f;
    tree[k*2].w+=tree[k].f*(tree[k*2].r-tree[k*2].l+1);
    tree[k*2+1].w+=tree[k].f*(tree[k*2+1].r-tree[k*2+1].l+1);
    tree[k].f=0;
}
void ask(int k)//单点查询
{
    if(tree[k].l==tree[k].r)
    {
        ans=tree[k].w;
        return;
    }
    if(tree[k].f) down(k);
    int m=(tree[k].l+tree[k].r)/2;
    if(x<=m) ask(k*2);
    else ask(k*2+1);
}
void ask_interval(int k)//区间查询
{
    if(tree[k].l>=a&&tree[k].r<=b)
    {
        ans+=tree[k].w;
        return;
    }
    if(tree[k].f) down(k);
    int m=(tree[k].l+tree[k].r)/2;
    if(a<=m) ask_interval(k*2);
    if(b>m) ask_interval(k*2+1);
}
void change_interval(int k)//区间修改
{
    if(tree[k].l>=a&&tree[k].r<=b)
    {
        tree[k].w+=(tree[k].r-tree[k].l+1)*y;
        tree[k].f+=y;
        return;
    }
    if(tree[k].f) down(k);
    int m=(tree[k].l+tree[k].r)/2;
    if(a<=m) change_interval(k*2);
    if(b>m) change_interval(k*2+1);
    tree[k].w=tree[k*2].w+tree[k*2+1].w;
}
int main()
{
    int n,m;
    char s;
    scanf("%d%d",&n,&m);
    build(1,1,n);
    while(m--)
    {
        scanf(" %c",&s);
        if(s=='Q')
        {
            scanf("%d%d",&a,&b);
            ans=0;
            ask_interval(1);
            printf("%lld\n",ans);
            if(m==0)
                break;

        }
        if(s=='C')
        {
            scanf("%d%d%d",&a,&b,&y);
            change_interval(1);
        }
    }
    return 0;
}