A Simple Problem with Integers（线段树+区间修改+区间询问模板）

（POJ - 3468）A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072K

Total Submissions: 119060 Accepted: 36974

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.

题目大意：一个有n个数的序列支持两种操作：C：对区间[l,r]中的每一个数都加一个值val；Q：询问区间[l,r]的和。

思路：线段树模板。区间修改，区间询问。

代码：

#include<cstdio>#define ls o<<1#define lr o<<1|1using namespace std;typedef long long LL;const int maxn=100005;LL a[maxn];int n,m;struct node{    int l,r;    LL sum;    LL lazy;    int len;}T[maxn<<2];void pushup(int o){    T[o].sum=T[ls].sum+T[lr].sum;}void build(int o,int l,int r){    T[o].l=l;    T[o].r=r;    T[o].lazy=0;//设置延迟标记     T[o].len=r-l+1;    if(l==r)//叶子节点     {        T[o].sum=a[l];        return;    }    int mid=(l+r)>>1;    build(ls,l,mid);//递归构造左子树     build(lr,mid+1,r);//递归构造右子树     pushup(o);//根据左右子树节点的值向上更新当前结点的值 }/*当前结点的标志向孩子结点传递*/ void pushdown(int o){     T[ls].sum+=T[o].lazy*T[ls].len;    T[lr].sum+=T[o].lazy*T[lr].len;    T[ls].lazy+=T[o].lazy;    T[lr].lazy+=T[o].lazy;     T[o].lazy=0;//传递后，当前结点标记清空 }void update(int o,int l,int r,int lazy){    if(T[o].l==l&&T[o].r==r)    {        T[o].sum+=T[o].len*lazy;        T[o].lazy+=lazy;        return;    }    if(T[o].lazy) pushdown(o);    int mid=(T[o].l+T[o].r)>>1;    if(mid>=r) update(ls,l,r,lazy);    else if(mid<l) update(lr,l,r,lazy);    else    {        update(ls,l,mid,lazy);        update(lr,mid+1,r,lazy);    }    //根据左右子树的值回溯向上更新当前结点的值     pushup(o);}LL query(int o,int l,int r){    if(T[o].l==l&&T[o].r==r) return T[o].sum;    if(T[o].lazy) pushdown(o);    int mid=(T[o].l+T[o].r)>>1;    if(mid>=r) return query(ls,l,r);    else if(mid<l) return query(lr,l,r);    else return query(ls,l,mid)+query(lr,mid+1,r);}int main(){    scanf("%d%d",&n,&m);    for(int i=1;i<=n;i++) scanf("%lld",a+i);    build(1,1,n);    while(m--)    {        char op[3];        scanf("%s",op);        if(op[0]=='C')        {            int l,r,v;            scanf("%d%d%d",&l,&r,&v);            update(1,l,r,v);        }        else        {            int l,r;            scanf("%d%d",&l,&r);            printf("%lld\n",query(1,l,r));        }    }    return 0;}