A Simple Problem with Integers (线段树区间更新）

这题是上一次的扩展有关区间更新的问题，不能用for循环的，因为每次要更新的，复杂度变高

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.

#7691097 zhuyinghui's solution for [Problem B]

StatusAcceptedTime2547msMemory7248kBLength2179LangG++Submitted2016-12-08 14:17:10Shared

Select Code

#include <iostream>#include <cstdio>#include <cstring>#define maxn 100010using namespace std;struct node{    int l,r;    long long val,add;}tree[maxn<<2];int a[maxn];long long sum;void pushup(int rt){    tree[rt].val=tree[rt<<1].val+tree[rt<<1|1].val;}void pushdown(int rt){    if(tree[rt].add)    {        tree[rt<<1].add+=tree[rt].add;        tree[rt<<1|1].add+=tree[rt].add;        tree[rt<<1].val+=tree[rt].add*(tree[rt<<1].r-tree[rt<<1].l+1);        tree[rt<<1|1].val+=tree[rt].add*(tree[rt<<1|1].r-tree[rt<<1|1].l+1);  //务必记得这些都是+=而不是等于这是个不容易发现的小错误        tree[rt].add=0;    }}void build(int rt,int p,int q){    tree[rt].l=p;    tree[rt].r=q;    tree[rt].add=0;    if(p==q)    {        tree[rt].val=a[p];        return ;    }    build(rt<<1,p,(p+q)>>1);    build(rt<<1|1,(p+q)/2+1,q);    pushup(rt);    return ;}void update(int rt,int p,int q,int c){    if(p<=tree[rt].l&&tree[rt].r<=q)    {        tree[rt].add+=c;        tree[rt].val+=c*(tree[rt].r-tree[rt].l+1);        return ;    }    pushdown(rt);    int mid=(tree[rt].r+tree[rt].l)>>1;    if(p<=mid)    {        update(rt<<1,p,q,c);    }    if(q>mid)    {        update(rt<<1|1,p,q,c);    }    pushup(rt);}void query(int rt,int p,int q){    int mid=(tree[rt].r+tree[rt].l)>>1;    if(p==tree[rt].l&&q==tree[rt].r)    {        sum+=tree[rt].val;    }    else{    pushdown(rt);//此处务必加上这个因为在更新的时候并没有全部更新叶子节点有遗漏，询问的时候需要再遍历一遍    if(q<=mid)    {        query(rt<<1,p,q);    }    else if(p>mid)    {        query(rt<<1|1,p,q);    }    else    {        query(rt<<1,p,mid);        query(rt<<1|1,mid+1,q);    }    }    return ;}int main(){    int n,t,x,y,z;    scanf("%d %d",&n,&t);    for(int i=1;i<=n;i++)    {        scanf("%d",&a[i]);    }    build(1,1,n);    char ch;    while(t--)    {        getchar();        scanf("%c",&ch);        if(ch=='C')        {            scanf("%d %d %d",&x,&y,&z);            update(1,x,y,z);        }        if(ch=='Q')        {            sum=0;            scanf("%d %d",&x,&y);            query(1,x,y);            printf("%I64d\n",sum);        }    }    return 0;}