Pku oj 3468 A Simple Problem with Integers（线段树Lazy标记）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 95170 Accepted: 29652Case 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

不用lazy标记会超时

#include<stdio.h>#include<iostream>>#include<algorithm>#define N 100050using namespace std;typedef long long int ll;struct Node{    int left,right;    ll sum;    ll mark;}tree[N<<2];int num[N];void create(int root, int left, int right){    tree[root].left = left;    tree[root].right = right;    tree[root].mark = 0;    if(left == right)    {        tree[root].sum = num[left];        return ;    }    int mid = (left + right) >> 1;    create(root << 1, left, mid);    create(root << 1 | 1, mid + 1, right);    tree[root].sum = tree[root<<1].sum + tree[root<<1|1].sum;}void update(int root, int left, int right,int val){    if(tree[root].left == left && tree[root].right == right)    {        tree[root].mark += val;        return ;    }    tree[root].sum += (ll)val * (right - left + 1);    int mid = (tree[root].left + tree[root].right) >> 1;    if(right <= mid)        update(root << 1, left, right,val);    else if(left > mid)        update(root << 1 | 1, left, right,val);    else    {        update(root << 1, left, mid, val);        update(root << 1 | 1, mid + 1, right, val);    }}ll query(int root, int left, int right){    if(tree[root].left == left && tree[root].right == right)    {        return tree[root].sum + tree[root].mark * (ll)(right-left+1);    }    if(tree[root].mark)    {        tree[root<<1].mark += tree[root].mark;        tree[root<<1|1].mark += tree[root].mark;        tree[root].sum += (ll)(tree[root].right - tree[root].left + 1) * tree[root].mark;        tree[root].mark = 0;    }    int mid = (tree[root].left + tree[root].right)>>1;    if(right <= mid)        return query(root << 1, left, right);    else if(left > mid)        return query(root << 1 | 1, left, right);    else    {        return query(root << 1, left, mid) + query(root << 1 | 1, mid + 1, right);    }}int main(){    int n,q,cnt = 0;    char str[8];    while(~scanf("%d%d",&n,&q))    {        for(int i = 1; i <= n; i++)            scanf("%d", &num[i]);        create(1, 1, n);        for(int i = 1;i <= q; i++)        {            int x,y,z;            scanf("%s",&str);            if(str[0] == 'Q')            {                scanf("%d%d",&x,&y);                printf("%lld\n",query(1,x,y));            }            if(str[0] == 'C')            {                scanf("%d%d%d",&x,&y,&z);                update(1,x,y,z);            }        }    }    return 0;}