POJ 3468 A Simple Problem with Integers线段树成段更新

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题目链接：POJ3468

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 90092		Accepted: 28052
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

Hint

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

题意：n个数q次操作，Q代表求l到r的和，C代表从l到r每项各加上c。

题目分析：线段树的整段查询，hint里说貌似int不够用，这里注意改称long long后懒惰数组的值也要long long型，连续加法或连续减法也可能超int。还有这里判断pushdown需要让懒惰数组不等于0，因为有可能出现负数。输入字符用％s比％c要快（很多），原因未知。

//
//  main.cpp
//  POJ3468
//
//  Created by teddywang on 16/5/24.
//  Copyright © 2016年 teddywang. All rights reserved.
//

#include <iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define N 100010
#define ll long long
ll sum[N<<2],lazy[N<<2];
int n,m;

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

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

void pushdown(int rt,int m)
{
    if(lazy[rt]!=0)
    {
        lazy[rt<<1]+=lazy[rt];
        lazy[rt<<1|1]+=lazy[rt];
        sum[rt<<1]+=lazy[rt]*(m-(m>>1));
        sum[rt<<1|1]+=lazy[rt]*(m>>1);
        lazy[rt]=0;
    }
}

void update(int v,int l1,int r1,int l,int r,int rt)
{
    if(l1<=l&&r<=r1)
    {
        sum[rt]+=(ll)v*(r-l+1);
        lazy[rt]+=v;
        return ;
    }
    pushdown(rt,r-l+1);
    int m=(l+r)>>1;
    if(l1<=m) update(v,l1,r1,lson);
    if(r1>m) update(v,l1,r1,rson);
    pushup(rt);
    
}

ll query(int l1,int r1,int l,int r,int rt)
{
    if(l1<=l&&r<=r1)
        return sum[rt];
    int m=(l+r)>>1;
    pushdown(rt,r-l+1);
    ll ans=0;
    if(l1<=m) ans+=query(l1,r1,lson);
    if(r1>m) ans+=query(l1,r1,rson);
    pushup(rt);
    return ans;
}

int main()
{
    while(cin>>n>>m)
    {
        build(1,n,1);
        for(int i=0;i<m;i++)
        {
            char c;
            cin>>c;
            if(c=='Q')
            {
                int l1,r1;
                scanf("%d%d",&l1,&r1);
                printf("%lld\n",query(l1,r1,1,n,1));
            }
            else
            {
                int l1,r1,v;
                scanf("%d%d%d",&l1,&r1,&v);
                update(v,l1,r1,1,n,1);
            }
        }
    }
}