POJ3468_A Simple Problem with Integers(線段樹/成段更新)
解題報告
題意：
略
思路：
線段樹成段更新，區間求和。
#include
#include
#include
#define LL long long
#define int_now int l,int r,int root
using namespace std;
LL sum[500000],lazy[500000];
void push_up(int root,int l,int r)
{
sum[root]=sum[root*2]+sum[root*2+1] + lazy[root]*(r-l+1);
}
void update(int root,int l,int r,int ql,int qr,LL v)
{
if(ql>r||qrr||qr
A Simple Problem with Integers
Time Limit: 5000MS
Memory Limit: 131072K
Total Submissions: 60817
Accepted: 18545
Case Time Limit: 2000MS
Description
You have N integers, A1, A2, ... , AN. 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, ... , AN. -1000000000 ≤ Ai ≤ 1000000000.
Each of the next Q lines represents an operation.
"C a b c" means adding c to each of Aa, Aa+1, ... , Ab. -10000 ≤ c ≤ 10000.
"Q a b" means querying the sum of Aa, Aa+1, ... , Ab.
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.
