POJ 3468 A Simple Problem with Integers（线段树区间更新模板题）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 108110 Accepted: 33683Case 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 51 2 3 4 5 6 7 8 9 10Q 4 4Q 1 10Q 2 4C 3 6 3Q 2 4

Sample Output

455915

线段树区间更新的模板题，存一下模板，lazy表示延迟更新，如果这一点更新不到，那么他的子节点就不用再更新了；

#include <iostream>#include <iomanip>#include<stdio.h>#include<string.h>#include<stack>#include<stdlib.h>#include<queue>#include<map>#include<math.h>#include<algorithm>#include<vector>#define maxn 100010#define ll long longusing namespace std;struct node{    ll v,L,R;    ll mid()    {        return (L+R)>>1;    }}t[maxn<<2];ll lazy[maxn<<2];void build(ll l,ll r,ll rt){    t[rt].L=l;    t[rt].R=r;    if(l==r)    {        scanf("%lld",&t[rt].v);        return;    }    ll m=t[rt].mid();    build(l,m,rt<<1);    build(m+1,r,rt<<1|1);    t[rt].v=t[rt<<1].v+t[rt<<1|1].v;}void pushdown(ll rt,ll m){     if(lazy[rt])     {         lazy[rt<<1]+=lazy[rt];         lazy[rt<<1|1]+=lazy[rt];         t[rt<<1].v+=lazy[rt]*(m-(m>>1));         t[rt<<1|1].v+=lazy[rt]*(m>>1);         lazy[rt]=0;     }}void update(ll num,ll l,ll r,ll rt){    if(t[rt].L==l&&t[rt].R==r)    {        lazy[rt]+=num;        t[rt].v+=num*(t[rt].R-t[rt].L+1);        return;    }    pushdown(rt,t[rt].R-t[rt].L+1);    ll m=t[rt].mid();    if(r<=m)        update(num,l,r,rt<<1);    else if(l>m)        update(num,l,r,rt<<1|1);    else    {        update(num,l,m,rt<<1);        update(num,m+1,r,rt<<1|1);    }    t[rt].v=t[rt<<1].v+t[rt<<1|1].v;}ll query(ll l,ll r,ll rt){    if(t[rt].L==l&&t[rt].R==r)    return t[rt].v;    pushdown(rt,t[rt].R-t[rt].L+1);    ll m=t[rt].mid();    if(r<=m) return query(l,r,rt<<1);    else if(l>m)  return query(l,r,rt<<1|1);    else        return query(l,m,rt<<1)+query(m+1,r,rt<<1|1);}int main(){    ll n,m,x,y,z;    scanf("%lld%lld",&n,&m);    build(1,n,1);    char op[5];    for(ll i=1;i<=m;i++)    {        scanf("%s",op);        if(op[0]=='C')        {            scanf("%lld%lld%lld",&x,&y,&z);            update(z,x,y,1);        }        else        {            scanf("%lld%lld",&x,&y);            printf("%lld\n",query(x,y,1));        }    }    return 0;}