POJ 3468 A Simple Problem with Integers 線段樹 區間修改

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 89433		Accepted: 27825
Case 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.

Source

POJ Monthly--2007.11.25, Yang Yi

點擊打開題目鏈接

想要寫點什麼，又不知怎麼寫。（ps：剛剛入門）

（藍橋，省賽，天梯。。。一堆的比賽就要將自己的小身板壓垮了，堅持！想想兩個月後就要加入了程序猿的隊伍，期待又膽怯。

越努力，越幸運---致默默爲祖國IT事業做出自己貢獻的網絡搬運工）

#include <cstdio>
#include <cstring>
#include <iostream>
using namespace std;
typedef long long LL;
#define Lson 2 * o, L, M			//左兒子 宏定義 會有意想不到的好處 
#define Rson 2 * o + 1, M + 1, R	//右兒子 
#define lo 2 * o
#define ro 2 * o + 1

const int maxn = 4 * 100000 + 10;

int ql, qr, v;
long long sum[maxn], add[maxn];
//向上傳遞，由 o 的左右子樹更新 o 
void pushUp(int o)
{
	sum[o] = sum[lo] + sum[ro];
}
//向下傳遞，維護節點o 
void pushDown(int o, int m)
{
	if (add[o])		//本節點有標記才傳遞 
	{
		add[lo] += add[o];	//左子樹點增加 
		add[ro] += add[o];	//右子樹點增加 
		sum[lo] += add[o] * (m - (m / 2));	//左子樹區間段增加 
		sum[ro] += add[o] * (m / 2);		//右子樹區間段增加 
		add[o] = 0;					//清除本節點標記 
	}
}
//建樹 
void Build(int o, int L, int R)
{
	add[o] = 0;						//建樹時初始節點都未標記 
	if (L == R) scanf("%lld", &sum[o]);	//葉子節點 
	else
	{
		int M = (L + R) / 2;
		Build(Lson);		//建立左子樹 
		Build(Rson);		//建立右子樹 
		pushUp(o);			//更新本節點 
	}
}
//更新區間 [ql,qr] 內的值加上 v 
void Update(int o, int L, int R)
{
	if (ql <= L && qr >= R) 	//遞歸邊界 
	{
		add[o] += v;			//累加邊界的add值 
		sum[o] += v * (R - L + 1);	//計算區間和 
	}
	else 
	{
		pushDown(o, R - L + 1);
		int M = (L + R) / 2;
		if (ql <= M) Update(Lson);   //先遞歸更新左子樹或者右子樹  
		if (qr > M) Update(Rson);
		pushUp(o);				//更新本節點 
	}
}
//查詢區間[ql,qr] 的和 
LL Query(int o, int L, int R)
{
	if (ql <= L && qr >= R) 	//遞歸邊界 
	{
		return sum[o];
	}
	else
	{
		pushDown(o, R - L + 1);	//用邊界區間的附加信息更新答案 
		int M = (L + R) / 2;
		LL ans = 0;				//遞歸統計，累加參數add 
		if (ql <= M) ans += Query(Lson);
		if (qr > M) ans += Query(Rson);
		return ans;
	}
}

int main()
{
	int n, q;
	while (~scanf("%d%d", &n, &q))
	{
		Build(1, 1, n);
	
		char ch[5];
		for (int i = 1; i <= q; i++)
		{
			scanf("%s", ch);
			if (ch[0] == 'Q')
			{
				scanf("%d%d", &ql, &qr);
				printf("%lld\n", Query(1, 1, n));
			}
			else
			{
				scanf("%d%d%d", &ql, &qr, &v);
				Update(1, 1, n);
			}
		}
	}
	return 0;
}