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

题目链接：http://poj.org/problem?id=3468

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 65789 Accepted: 20259Case 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

Hint

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

Source

POJ Monthly--2007.11.25, Yang Yi

线段树成段更新 模板题就不多说了

#include<cstdio>#include<cstring>#include<cmath>#include<cstdlib>#include<iostream>#include<algorithm>#include<sstream>#include<vector>#include<map>#include<list>#include<set>#include<queue>#define LL long long#define lson l,m,rt<<1#define rson m+1,r,rt<<1 | 1using namespace std;const int maxn=100005;LL add[maxn<<2],sum[maxn<<2];void PushUp(int rt){    sum[rt]=sum[rt<<1]+sum[rt<<1|1];}void PushDown(int rt,int m){    if(add[rt])    {        add[rt<<1]+=add[rt];        add[rt<<1|1]+=add[rt];        sum[rt<<1]+=(m-(m>>1))*add[rt];        sum[rt<<1|1]+=(m>>1)*add[rt];        add[rt]=0;    }}void build(int l,int r,int rt){    add[rt]=0;    if(l==r)    {        scanf("%lld",&sum[rt]);        return ;    }    int m=(l+r)>>1;    build(lson);    build(rson);    PushUp(rt);}void update(int L,int R,int x,int l,int r,int rt){    if(L<=l&&r<=R)    {        sum[rt]+=(LL)((r-l+1)*x);        add[rt]+=(LL)x;        return ;    }    PushDown(rt,r-l+1);    int m=(l+r)>>1;    if(L<=m) update(L,R,x,lson);    if(R>m) update(L,R,x,rson);    PushUp(rt);}LL query(int L,int R,int l,int r,int rt){    if(L<=l&&r<=R)    {        return sum[rt];    }    PushDown(rt,r-l+1);    int m=(l+r)>>1;    LL ans=0;    if(L<=m) ans+=query(L,R,lson);    if(R>m) ans+=query(L,R,rson);    return ans;}int main(){    int n,m,x,y,v;    char op[5];    while(~scanf("%d%d",&n,&m))    {        build(1,n,1);        for(int i=0;i<m;i++)        {            scanf("%s",op);            if(op[0]=='Q') scanf("%d%d",&x,&y),printf("%lld\n",query(x,y,1,n,1));            else scanf("%d%d%d",&x,&y,&v),update(x,y,v,1,n,1);        }    }    return 0;}