A Simple Problem with Integers（区间更新）

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 121236 Accepted: 37645Case 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.

线段树区间更新，需要用到延迟标记，需要更新的区间才会改变值，对下面停止更新。

#include<cstdio>#include<cmath>#include<cstring>#include<iostream>#include<algorithm>using namespace std;typedef long long ll;const int N=1e5+10;#define lson l,m,rt<<1#define rson m+1,r,rt<<1|1ll sum[N<<2];ll add[N<<2];// 记录节点每个数值该增加多少struct Node{    int l,r;    int mid(){    return (l+r)>>1;    }}tree[N<<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]+=add[rt]*(m-(m>>1));        sum[rt<<1|1]+=add[rt]*(m>>1);        add[rt]=0;    }}void build(int l,int r,int rt){    tree[rt].l=l;    tree[rt].r=r;    add[rt]=0;    if(l==r)    {        scanf("%I64d",&sum[rt]);        return;    }    int m=tree[rt].mid();    build(lson);    build(rson);    PushUp(rt);}void updata(int c,int l,int r,int rt){    if(tree[rt].l==l&&tree[rt].r==r)    {        add[rt]+=c;        sum[rt]+=(ll)c*(r-l+1);        return;    }    if(tree[rt].l==tree[rt].r) return;    PushDown(rt,tree[rt].r-tree[rt].l+1);    int m=tree[rt].mid();    if(r<=m) updata(c,l,r,rt<<1);    else if(l>m) updata(c,l,r,rt<<1|1);    else {        updata(c,l,m,rt<<1);        updata(c,m+1,r,rt<<1|1);    }    PushUp(rt);}ll query(int l,int r,int rt){    if(l==tree[rt].l&&r==tree[rt].r) return sum[rt];    PushDown(rt,tree[rt].r-tree[rt].l+1);    int m=tree[rt].mid();    ll res=0;    if(r<=m) res+=query(l,r,rt<<1);    else if(l>m) res+=query(l,r,rt<<1|1);    else    {        res+=query(l,m,rt<<1);        res+=query(m+1,r,rt<<1|1);    }    return res;}int main(){    int n,m;;    while(~scanf("%d%d",&n,&m))    {        build(1,n,1);        while(m--)        {            char ch[2];            scanf("%s",ch);            int a,b,c;            if(ch[0]=='Q')            {                scanf("%d%d",&a,&b);                printf("%I64d\n",query(a,b,1));            }            else            {                scanf("%d%d%d",&a,&b,&c);                updata(c,a,b,1);            }        }    }}

其实不太懂，就当先撂这个模板吧