POJ 3468 A Simple Problem with Integers

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 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

Hint

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

题目链接

题意：区间修改与查询区间和

分析

线段树的基本操作，使用lazy标记的思想。

#include<iostream>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#include<algorithm>
#include<cstring>
#include <queue>
#include <vector>
#include<bitset>
#include<map>
#include<deque>
using namespace std;
typedef long long LL;
const int maxn = 1e5+150;
const int mod = 77200211+233;
typedef pair<int,int> pii;
#define X first
#define Y second
#define pb push_back
#define mp make_pair
#define ms(a,b) memset(a,b,sizeof(a))
const int inf = 0x3f3f3f3f;
#define lson l,m,2*rt
#define rson m+1,r,2*rt+1

LL sum[maxn<<2],lazy[maxn<<2];

struct node{
    int l,r;
    int mid(){
        return (l+r)>>1;
    }
}tree[maxn<<2];
//更新父亲信息
void PushUp(int rt){
    sum[rt]=sum[rt<<1]+sum[rt<<1|1];
}
//更新子孙信息
void PushDown(int rt,int len){
    if(lazy[rt]){
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        sum[rt<<1] += lazy[rt]*(len-len/2);
        sum[rt<<1|1] += lazy[rt]*(len/2);
        lazy[rt]=0;
    }
}

void build(int l,int r,int rt){
    tree[rt].l=l,tree[rt].r=r;
    lazy[rt]=0;
    if(l==r){
        scanf("%lld",&sum[rt]);
        return;
    }
    int m=tree[rt].mid();
    build(lson);
    build(rson);
    PushUp(rt);
}

void update(int d,int l,int r,int rt){
    if(tree[rt].l==l&&tree[rt].r==r){
        lazy[rt] += d; //打上标记
        sum[rt] += d*(r-l+1);
        return;
    }
    if(tree[rt].l==tree[rt].r) return;

    PushDown(rt,tree[rt].r-tree[rt].l+1);

    int m=tree[rt].mid();
    if(r<=m) update(d,l,r,rt<<1);
    else if(l>m) update(d,l,r,rt<<1|1);
    else{
        update(d,l,m,rt<<1);
        update(d,m+1,r,rt<<1|1);
    }
    PushUp(rt);
}

LL query(int l,int r,int rt){
    if(tree[rt].l==l&&tree[rt].r==r){
        return sum[rt];
    }
    PushDown(rt,tree[rt].r-tree[rt].l+1);
    int m = tree[rt].mid();
    LL res=0;
    if(r<=m){
        res+=query(l,r,rt<<1);
    }else if(l>m){
        res+=query(l,r,rt<<1|1);
    }else{
        res+=query(l,m,rt<<1);
        res+=query(m+1,r,rt<<1|1);
    }
    return res;
}

int main(){
    int n,m;
    while(~scanf("%d%d",&n,&m)){
        build(1,n,1);
        while(m--){
            char ch[2];
            scanf("%s",ch);
            int a,b,c;
            if(ch[0]=='Q'){
                scanf("%d%d",&a,&b);
                cout<<query(a,b,1)<<endl;
            }else{
                scanf("%d%d%d",&a,&b,&c);
                update(c,a,b,1);
            }

        }

    }
    return 0;
}