线段树 poj 3468 A Simple Problem with Integers 区间add更新

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 41899 Accepted: 12169Case 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.

#include<iostream>#include<cstdio>#include<cmath>#include<cstring>#include<algorithm>#define N 222222using namespace std;int num[N];struct Tree{    int l;    int r;    long long sum;    long long col;} tree[N*4];void push_up(int rt){    tree[rt].sum=tree[rt<<1].sum+tree[rt<<1|1].sum;}void push_down(int rt,int m){    if (tree[rt].col!=0)    {        tree[rt<<1].col+=tree[rt].col;        tree[rt<<1|1].col+=tree[rt].col;        tree[rt<<1].sum+=(long long)(m-(m/2))*tree[rt].col;        tree[rt<<1|1].sum+=(long long)(m/2)*tree[rt].col;        tree[rt].col=0;    }}void build(int root,int l,int r){    tree[root].l=l;    tree[root].r=r;    tree[root].col=0;    if(tree[root].l==tree[root].r)    {        tree[root].sum=num[l];        return;    }    int mid=(l+r)/2;    build(root<<1,l,mid);    build(root<<1|1,mid+1,r);    push_up(root);}void update(int root,int L,int R,int val){    if(L<=tree[root].l&&R>=tree[root].r)    {        tree[root].col+=val;        tree[root].sum+=(long long)val*(tree[root].r-tree[root].l+1);        return;    }    push_down(root,tree[root].r-tree[root].l+1);    int mid=(tree[root].l+tree[root].r)/2;    if (L<=mid)        update(root<<1,L,R,val);    if (R>mid)        update(root<<1|1,L,R,val);    push_up(root);}long long query(int root,int L,int R){    if(L<=tree[root].l&&R>=tree[root].r)        return tree[root].sum;    push_down(root,tree[root].r-tree[root].l+1);    int mid=(tree[root].l+ tree[root].r)/2;    long long ret=0;    if(L<=mid) ret+=query(root<<1,L,R);    if(R>mid) ret+=query(root<<1|1,L,R);    return ret;}int main(){    int n,q;    char s[2];    scanf("%d%d",&n,&q);    for (int i=1; i<=n; i++) scanf("%d",&num[i]);    build(1,1,n);    while (q--)    {        int x,y,z;        scanf("%s",s);        if (s[0]=='Q')        {            scanf("%d%d",&x,&y);            printf("%I64d\n",query(1,x,y));        }        else if (s[0]=='C')        {            scanf("%d%d%d",&x,&y,&z);            update(1,x,y,z);        }    }    return 0;}