POJ 3468 A Simple Problem with Integers （线段树【区间更新】）

A Simple Problem with Integers

Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 96409 Accepted: 30061Case 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 <cstdio>#include <cstring>using namespace std;const int maxn = 100000 + 10;typedef long long ll;int n, q;char ch;ll a, b, c;ll s[maxn];ll anssum;struct node{    ll l, r;    ll addv, sum;}tree[maxn << 2];void maintain(int id){    if (tree[id].l >= tree[id].r)        return;    tree[id].sum = tree[id << 1].sum + tree[id << 1 | 1].sum;}void pushdown(int id){    if (tree[id].l >= tree[id].r)        return;    if (tree[id].addv){        int tmp = tree[id].addv;        tree[id << 1].addv += tmp;        tree[id << 1 | 1].addv += tmp;        tree[id << 1].sum += (tree[id << 1].r - tree[id << 1].l + 1) * tmp;        tree[id << 1 | 1].sum += (tree[id << 1 | 1].r - tree[id << 1 | 1].l + 1) * tmp;        tree[id].addv = 0;    }}void build(int id, ll l, ll r){    tree[id].l = l;    tree[id].r = r;    tree[id].addv = 0;    tree[id].sum = 0;    if (l == r){        tree[id].sum = s[l];        return;    }    ll mid = (l + r) >> 1;    build(id << 1, l, mid);    build(id << 1 | 1, mid + 1, r);    maintain(id);}void updateAdd(int id, ll l, ll r, ll val){    if (tree[id].l >= l && tree[id].r <= r){        tree[id].addv += val;        tree[id].sum += (tree[id].r - tree[id].l  + 1) * val;        return;    }    pushdown(id);    ll mid = (tree[id].l + tree[id].r) >> 1;    if (l <= mid)        updateAdd(id << 1, l, r, val);    if (mid < r)        updateAdd(id << 1 | 1, l, r, val);    maintain(id);}void query(int id, ll l, ll r){    if (tree[id].l >= l && tree[id].r <= r){        anssum += tree[id].sum;        return;    }    pushdown(id);    ll mid = (tree[id].l + tree[id].r) >> 1;    if (l <= mid)        query(id << 1, l, r);    if (mid < r)        query(id << 1 | 1, l, r);    maintain(id);}int main(){    while (scanf("%d%d", &n, &q) != EOF){        for (int i = 1; i <= n; i++){            scanf("%lld", &s[i]);        }        build(1, 1, n);        while (q--){            scanf("%*c%c", &ch);            if (ch == 'C'){                scanf("%lld%lld%lld", &a, &b, &c);                updateAdd(1, a, b, c);            }            else{                scanf("%lld%lld", &a, &b);                anssum = 0;                query(1, a, b);                printf("%lld\n", anssum);            }        }    }    return 0;}