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.
应该算是板子题，主要就是加上懒标记，加快但是我在poj上1998ms，在百炼oj上607ms不知为何，
AC code：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cstring>

using namespace std;
typedef long long ll;

const int maxn = 1e5+5;

struct segtree{
    int l,r,lazy;
    ll sum,val;
}ss[maxn<<2];
int n,q;

void pushDown(int rt){
    if(ss[rt].lazy) {
        ss[rt<<1].lazy = ss[rt<<1|1].lazy = 1;
        ss[rt<<1].val += ss[rt].val;
        ss[rt<<1|1].val += ss[rt].val;
        ss[rt<<1].sum += (ss[rt<<1].r-ss[rt<<1].l + 1) * ss[rt].val;
        ss[rt<<1|1].sum += (ss[rt<<1|1].r-ss[rt<<1|1].l + 1) * ss[rt].val;
        ss[rt].lazy = 0; ss[rt].val = 0;
    }
}

void pushUp(int rt){
    ss[rt].sum = ss[rt<<1].sum + ss[rt<<1|1].sum;
}

void build(int l,int r,int rt){
    ss[rt].l = l,ss[rt].r = r;
    if ( l == r ) {
        scanf("%lld",&ss[rt].sum);ss[rt].val = 0;ss[rt].lazy = 0;
        return;
    }
    int mid = (l+r)>>1;
    build(l,mid,rt<<1); build(mid+1,r,rt<<1|1);
    pushUp(rt); ss[rt].lazy = ss[rt].val = 0;
}

void update(int l,int r,int rt,int val){
    if( ss[rt].l == l && ss[rt].r == r) {
        ss[rt].sum += (ss[rt].r - ss[rt].l + 1) * val;
        ss[rt].lazy = 1, ss[rt].val += val;
        return;
    }
    pushDown(rt);
    int mid = (ss[rt].l+ss[rt].r)>>1;
    if ( r <= mid ) {
        update(l ,r ,rt<<1 ,val );
    } else if ( l > mid ){
        update(l ,r ,rt<<1|1 ,val );
    } else {
        update(l ,mid ,rt<<1 ,val );
        update(mid+1 ,r ,rt<<1|1 ,val );
    }
    pushUp(rt);
}

ll query(int l,int r,int rt){
    if (ss[rt].l == l && ss[rt].r == r  ) {
        return ss[rt].sum;
    }
    pushDown(rt);
    int mid = (ss[rt].l+ss[rt].r)>>1; ll sum = 0;
    if ( r <= mid ) {
        return sum + query(l,r,rt<<1);
    } else if ( l>mid ) {
        return sum  + query(l,r,rt<<1|1);
    } else {
        return sum + query(l,mid,rt<<1) + query(mid+1,r,rt<<1|1);
    }
}

int main(){
    char str;
    int a,b,c;
    while(~scanf("%d %d",&n,&q))
    {
        build(1,n,1);
        while(q--){
            getchar();
            scanf("%c",&str);
            if ( str == 'Q' ) {
                scanf("%d %d",&a,&b);
                if(a>b) { int t = a; a = b; b = t;  }
                printf("%lld\n",query(a,b,1));
            } else if ( str == 'C' ) {
                scanf("%d %d %d",&a,&b,&c);
                if(a>b) { int t = a; a = b; b = t;  }
                update(a,b,1,c);
            }
        }
    }
    return 0;
}