poj3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 71334 Accepted: 21998Case 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

仍然是一道线段树区间更新的问题，上次做的那道题目只涉及到更新，查询的时候只查询根节点，而这个也是需要查询的，同样的道理，查询一个节点之前要完成其之前未完成的向下更新操作，而且，由于题目给出的操作是在每一个节点上都加上一个数，但是在向下更新的时候，下面的节点也有可能存在更早的未完成的向下更新操作，而且这两次更新的效果是累加的，所以延时标志也应该是累加的。

#include <iostream>#include <stdio.h>#define MAX 800000using namespace std;struct Tree{    long long sum;    long long flag;};Tree tree[MAX];void PushUp(int rt){    tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;}void PushDown(int rt,int len){    if(tree[rt].flag!=0)    {        tree[rt<<1].flag+=tree[rt].flag;        tree[rt<<1|1].flag+=tree[rt].flag;        tree[rt<<1].sum+=(len-(len>>1))*tree[rt].flag;        tree[rt<<1|1].sum+=(len>>1)*tree[rt].flag;        tree[rt].flag=0;    }}void Build(int l,int r,int rt){    tree[rt].flag=0;    if(l==r)    {        cin>>tree[rt].sum;        return;    }    int m=(l+r)>>1;    Build(l,m,rt<<1);    Build(m+1,r,rt<<1|1);    PushUp(rt);}void Update(int L,int R,int d,int l,int r,int rt){    if(L<=l&&R>=r)    {        tree[rt].flag+=d;        tree[rt].sum+=(r-l+1)*d;        return;    }    PushDown(rt,r-l+1);    int m=(l+r)>>1;    if(L<=m)        Update(L,R,d,l,m,rt<<1);    if(R>m)        Update(L,R,d,m+1,r,rt<<1|1);    PushUp(rt);}long long Query(int L,int R,int l,int r,int rt){    if(L<=l&&R>=r)    {        return tree[rt].sum;    }    PushDown(rt,r-l+1);    int m=(l+r)>>1;    long long sum=0;    if(L<=m)        sum+=Query(L,R,l,m,rt<<1);    if(R>m)        sum+=Query(L,R,m+1,r,rt<<1|1);    return sum;}int main(){    int n,q;    int a,b,c;    int i;    char op;    scanf("%d%d",&n,&q);    {        Build(1,n,1);        while(q--)        {            cin>>op;            if(op=='Q')            {                scanf("%d%d",&a,&b);                cout<<Query(a,b,1,n,1)<<endl;            }            else if(op=='C')            {                scanf("%d%d%d",&a,&b,&c);                Update(a,b,c,1,n,1);            }        }    }    return 0;}