poj 3468 , 线段树

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 50205		Accepted: 14911
Case 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 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.

Source

POJ Monthly--2007.11.25, Yang Yi

 

线段树，区间和，延迟标记

 

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
#include<set>
#include<queue>
#include<string>
#include<cmath>
#include<fstream>
#include<iomanip>

using namespace std;

#define LL long long
#define lson rt<<1, l, m
#define rson rt<<1|1, m, r
#define MAXN 111111

int n, q;
LL sum[MAXN<<2], todo[MAXN<<2];

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

void push_down(int rt, int l, int r){
    if(!todo[rt]) return;
    todo[rt<<1] += todo[rt];    todo[rt<<1|1] += todo[rt];
    int m = l + r >> 1;
    sum[rt<<1] += (m - l) * todo[rt];
    sum[rt<<1|1] += (r - m) * todo[rt];
    todo[rt] = 0;
}

void build(int rt, int l, int r){
    todo[rt] = 0;
    if(l+1 == r){
        scanf(" %lld", sum+rt);  return;
    }
    int m = l + r >> 1;
    build(lson);
    build(rson);
    push_up(rt);
}

void update(int rt, int l, int r, int cl, int cr, int tc){
    if(cl<=l && cr>=r){
        todo[rt] += tc;  sum[rt] += (r - l) * tc;
        return;
    }
    int m = l + r >> 1;
    push_down(rt, l, r);
    if(cl < m) update(lson, cl, cr, tc);
    if(cr > m) update(rson, cl, cr, tc);
    push_up(rt);
}

LL query(int rt, int l, int r, int cl, int cr){
    if(cl<=l && cr>=r) return sum[rt];
    int m = l + r >> 1;
    LL ret = 0;
    push_down(rt, l, r);
    if(cl < m) ret += query(lson, cl, cr);
    if(cr > m) ret += query(rson, cl, cr);
    return ret;
}

int main(){
//    freopen("C:\\Users\\Administrator\\Desktop\\in.txt","r",stdin);
    while(scanf(" %d %d", &n, &q)==2){
        build(1, 1, 1+n);
        while(q--){
            int a, b, c;    char ch;
            scanf(" %c %d %d", &ch, &a, &b);    b++;
            if(ch == 'C'){
                scanf(" %d", &c);
                update(1, 1, 1+n, a, b, c);
            }
            else printf("%lld\n", query(1, 1, 1+n, a, b));
        }
    }
    return 0;
}