poj3468 A Simple Problem with Integers(线段树)

A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072KTotal Submissions: 114537 Accepted: 35529Case 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 5 1 2 3 4 5 6 7 8 9 10 Q 4 4 Q 1 10 Q 2 4 C 3 6 3 Q 2 4

Sample Output

4 55 9 15

Hint

The sums may exceed the range of 32-bit integers 思路 add这个数组相当于记了一个懒惰标记，只有在求和 和 查询的时候才会更新数值。其他就易于理解了 #include<iostream> #include<cmath> #include<cstring> #include<vector> #include<stdlib.h> #include<stdio.h> #include<algorithm> #include<map> #include <set> #include <list> #include <deque> #include<sstream> #include<time.h> #define pi 3.1415926 #define N 2005 #define M 15 #define INF 0x6f6f6f6f #define lson l , m , rt << 1 #define rson m + 1 , r , rt << 1 | 1 using namespace std; typedef long long ll; const int day =21252; const int mod=1000000007; const int maxn = 111111; ll add[maxn<<2]; ll sum[maxn<<2]; void PushUp(int rt) {//向上更新节点 sum[rt] = sum[rt<<1] + sum[rt<<1|1]; } void PushDown(int rt,int m) {//向下更新节点 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) {//建树 add[rt] = 0; if (l == r) { scanf("%lld",&sum[rt]); return; } int m = (l + r) >> 1; build(lson); build(rson); PushUp(rt); } void update(int L,int R,int c,int l,int r,int rt) {//更新 if (L <= l && r <= R) { add[rt] += c; sum[rt] += (ll)c * (r - l + 1); return ; } PushDown(rt , r - l + 1); int m = (l + r) >> 1; if (L <= m) update(L , R , c , lson); if (m < R) update(L , R , c , rson); PushUp(rt); } ll query(int L,int R,int l,int r,int rt) {//求和 if (L <= l && r <= R) { return sum[rt]; } PushDown(rt , r - l + 1); int m = (l + r) >> 1; ll ret = 0; if (L <= m) ret += query(L , R , lson); if (m < R) ret += query(L , R , rson); return ret; } int main() { int n,q; scanf("%d%d",&n,&q); build(1 , n , 1); while (q --) { char op[2]; int a , b , c; scanf("%s",op); if (op[0] == 'Q') { scanf("%d%d",&a,&b); printf("%lld\n",query(a , b , 1 , n , 1)); } else { scanf("%d%d%d",&a,&b,&c); update(a , b , c , 1 , n , 1); } } return 0; } .