POJ 3468(线段树模板 Lazy)

A Simple Problem with Integers
Time Limit: 5000MS Memory Limit: 131072K
Total Submissions: 111585 Accepted: 34748
Case Time Limit: 2000MS
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.
Source

POJ Monthly–2007.11.25, Yang Yi

#include <iostream>#include <cstring>#include <algorithm>#include <cstdio>#include <cmath>#define maxn 111111using namespace std;typedef long long ll;ll num[maxn];ll segtree[maxn*20];ll add[maxn*20];void init(int n){    for(int i=1;i<=n;i++)    {        segtree[i]=0;    }}void PushDown(int rt,int l){    if(add[rt]!=0)    {        add[rt<<1]+=add[rt];        add[rt<<1|1]+=add[rt];        segtree[rt<<1]+=(l-(l>>1))*add[rt];        segtree[rt<<1|1]+=(l>>1)*add[rt];        add[rt]=0;    }}void plant(ll left,ll right,ll rt){    add[rt]=0;    if(left==right)    {        scanf("%I64d",&segtree[rt]);        return ;    }    ll mid=(left+right)>>1;    plant(left,mid,rt<<1);    plant(mid+1,right,rt<<1|1);    segtree[rt]=segtree[rt<<1]+segtree[rt<<1|1];}ll query(ll left,ll right,ll rt,ll L,ll R){    if(L<=left&&right<=R)    {        return segtree[rt];    }    PushDown(rt,right-left+1);    ll mid=(left+right)/2;    ll ans=0;    if(L<=mid)        ans+=query(left,mid,rt<<1,L,R);    if(R>mid)        ans+=query(mid+1,right,rt<<1|1,L,R);    return ans;}void update(ll left,ll right,ll rt,ll l,ll r,ll up){    if(l<=left&&r>=right)    {        segtree[rt]+=(ll)up*(right-left+1);        add[rt]+=up;        return;    }    PushDown(rt,right-left+1);    int mid=(left+right)>>1;    if(l<=mid)update(left,mid,rt<<1,l,r,up);    if(r>mid)update(mid+1,right,rt<<1|1,l,r,up);    segtree[rt]=segtree[rt<<1]+segtree[rt<<1|1];}int main(){    int n,m;    while(scanf("%d%d",&n,&m)!=EOF)    {        memset(segtree,0,sizeof(segtree));        memset(add,0,sizeof(add));        plant(1,n,1);        ll a,b,c;        char op[10];        for(int i=0;i<m;i++)        {            scanf("%s",op);            if(op[0]=='Q')            {                scanf("%I64d%I64d",&a,&b);                printf("%I64d\n",query(1,n,1,a,b));            }            else            {                scanf("%I64d%I64d%I64d",&a,&b,&c);                update(1,n,1,a,b,c);            }        }    }    return 0;}