POJ 3468 线段树（区间更新）

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A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 85553 Accepted: 26573Case 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

Hint

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

线段树的区间更新，代码量比较大，写的时候容易出错。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;const int maxn = 100005 ;struct node{    int l,r;    long long sum,add ;};node tree[3*maxn] ;int N,q;void build(int i,int l,int r){    tree[i].l=l;    tree[i].r=r;    tree[i].add=0;    tree[i].sum=0;    if(l==r)return ;    int mid = (l+r)/2 ;    build(2*i,l,mid);    build(2*i+1,mid+1,r);}void pushdown(int i,int l,int r){    int mid = (l+r)/2 ;    tree[2*i].add+=tree[i].add ;    tree[2*i+1].add+=tree[i].add ;    tree[2*i].sum+=tree[i].add*(mid-l+1) ;    tree[2*i+1].sum+=tree[i].add*(r-mid) ;    tree[i].add=0 ;}void update(int i,int l,int r,int yl,int yr,long long v){    if(yl<=l&&r<=yr){        tree[i].add+=v;        tree[i].sum+=v*(r-l+1) ;    }else{        int mid=(l+r)/2 ;        pushdown(i,l,r) ;        if(yl<=mid)update(2*i,l,mid,yl,yr,v);        if(mid<yr)update(2*i+1,mid+1,r,yl,yr,v);        tree[i].sum=tree[2*i].sum+tree[2*i+1].sum ;    }}long long query(int i,int l,int r,int ql,int qr){    long long ret = 0 ;    if(ql<=l&&r<=qr){        return tree[i].sum ;    }else{        int mid = (l+r)/2 ;        pushdown(i,l,r) ;        if(ql<=mid)ret+=query(2*i,l,mid,ql,qr);        if(mid<qr)ret+=query(2*i+1,mid+1,r,ql,qr) ;        return ret ;    }}int main(){    while(~scanf("%d%d",&N,&q)){        memset(tree,0,sizeof(tree));        build(1,1,N);        for(int i=1;i<=N;i++){            int a ;            scanf("%d",&a);            update(1,1,N,i,i,a);        }        for(int i=0;i<q;i++){           int a,b,c;           char str[10] ;           scanf("%s",str);           if(str[0]=='C'){                scanf("%d %d %d",&a,&b,&c);                update(1,1,N,a,b,c);           }else{                scanf("%d%d",&a,&b);                printf("%I64d\n",query(1,1,N,a,b));           }        }    }    return 0;}

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