poj 3468 A Simple Problem with Integers（线段树，lazy思想 经典题）

A Simple Problem with Integers
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Time Limit:5000MS
Memory Limit:131072KB
64bit IO Format:%I64d & %I64u

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 abc" means adding c to each of Aa, Aa+1, ... ,
Ab. -10000 ≤ c ≤ 10000.

"Q ab" 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

题意：

给你n个数，m次操作，每次操作有Q/C两种，Q:求出所给区间的和，C:给所给区间的每个值都加上a;

思路：

线段树，用sets[]给每个区间做标记

My Code:

#include<iostream>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<string>
#include<algorithm>
#include<cstdlib>
#include<set>
#include<queue>
#include<stack>
#include<vector>
#include<map>

#define N 100010
#define Mod 10000007
#define lson l,mid,idx<<1
#define rson mid+1,r,idx<<1|1
#define lc idx<<1
#define rc idx<<1|1
const double EPS = 1e-11;
const double PI = acos(-1.0);
const double E=2.718281828;
typedef long long ll;

const int INF=1000010;

using namespace std;

ll tree[N<<2],sets[N<<2];

void push_up(int idx)
{
    tree[idx]=tree[lc]+tree[rc];
}

void build(int l,int r,int idx)
{
    sets[idx]=0;
    if(l==r)
    {
        scanf("%I64d",&tree[idx]);
        return ;
    }
    int mid=(l+r)>>1;
    build(lson);
    build(rson);
    push_up(idx);
}

void push_down(int idx,int m)///m为区间长度
{
    if(sets[idx])
    {
        sets[rc]+=sets[idx];
        sets[lc]+=sets[idx];
        tree[lc]+=sets[idx]*(m-m/2);
        tree[rc]+=sets[idx]*(m/2);
        sets[idx]=0;
    }
}

void update(int l,int r,int idx,int x,int y,ll a)
{
    if(x<=l&&y>=r)
    {
        sets[idx]+=a;
        tree[idx]+=(ll)(r-l+1)*a;
        return;
    }
    push_down(idx,r-l+1);//注意此函数不要忘了
    int mid=(l+r)>>1;
    if(x<=mid)
        update(lson,x,y,a);
    if(y>mid)
        update(rson,x,y,a);
    push_up(idx);
}

ll query(int l,int r,int idx,int x,int y)
{
    if(x<=l&&y>=r)
        return tree[idx];
    push_down(idx,r-l+1);
    int mid=(l+r)>>1;
    ll ans=0;
    if(x<=mid)
        ans+=query(lson,x,y);
    if(y>mid)
        ans+=query(rson,x,y);
    return ans;
}

int main()
{
    //freopen("in.txt","r",stdin);
    ll n,m;
    while(~scanf("%I64d%I64d",&n,&m))
    {
        build(1,n,1);
        char c;
        ll a;
        int x,y;
        getchar();
        while(m--)
        {
            scanf("%c%d%d",&c,&x,&y);
            if(c=='Q')
                printf("%I64d\n",query(1,n,1,x,y));
            else
            {
                scanf("%I64d",&a);
                update(1,n,1,x,y,a);
            }
            getchar();
        }
    }
    return 0;
}