poj 3468 A Simple Problem with Integers

A Simple Problem with Integers Time Limit: 5000MS Memory Limit: 131072K Total Submissions: 51293 Accepted: 15271 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. Source POJ Monthly--2007.11.25, Yang Yi

[Submit]   [Go Back]   [Status]   [Discuss]

我的解答

/*=============================================================================
#     FileName: 3468.cpp
#         Desc: poj 3468
#       Author: zhuting
#        Email: [email protected]
#     HomePage: my.oschina.net/locusxt
#      Version: 0.0.1
#    CreatTime: 2013-12-08 11:25:00
#   LastChange: 2013-12-08 11:25:00
#      History:
=============================================================================*/
#include <cstdio>
#include <cstdlib>
#include <string>
#include <cstring>
#include <algorithm>
#define maxn 100005

class node/*区间结点*/
{
	public:
		int l, r;
		long long sum_, toadd;/*toadd为待加的值*/
};

int a[maxn] = {0};
node no[maxn << 2];
long long ans = 0;

void build(int le, int ri, int id)/*构建*/
{
	no[id].l = le;
	no[id].r = ri;
	no[id].toadd = 0;
	if (le == ri)/*区间中只有一个点*/
	{
		no[id].sum_ = a[le];
		return;
	}
	int mid = (le + ri) >> 1;
	build (le, mid, id << 1);/*递归建左右树*/
	build (mid + 1, ri, (id << 1) + 1);
	no[id].sum_ = no[id << 1].sum_ + no[(id << 1) + 1].sum_;
}

void ask(int le, int ri, int id)/*询问区间结点id,在[le, ri]区间的和*/
{
	if (no[id].l > ri || no[id].r < le)/*区间结点id不在该区间中*/
	{
		return;
	}
	if (no[id].l >= le && no[id].r <= ri)/*id被该区间完全覆盖,此时不需要计算待加值,直接加上sum_即可*/
	{
		ans += no[id].sum_;
		return;
	}
	/*id与该区间相交*/
	if (no[id].toadd)/*有未加项*/
	{
		/*更新两个子节点*/
		int tmp_id = id << 1;
		no[tmp_id].sum_ += (no[tmp_id].r - no[tmp_id].l + 1) * no[id].toadd;
		no[tmp_id].toadd += no[id].toadd;
		++tmp_id;
		no[tmp_id].sum_ += (no[tmp_id].r - no[tmp_id].l + 1) * no[id].toadd;
		no[tmp_id].toadd += no[id].toadd;
		no[id].toadd = 0;
	}
	/*然后再询问子节点*/
	int tmp_id = id << 1;
	ask(le, ri, tmp_id);
	ask(le, ri, tmp_id + 1);
	no[id].sum_ = no[tmp_id].sum_ + no[tmp_id + 1].sum_;
	return;
}

void update(int le, int ri, int id, long long delta)/*区间[le, ri]加上delta,对id的影响*/
{
	/*与上面询问类似*/
	if (no[id].l > ri || no[id].r < le)
		return;
	if (no[id].l >= le && no[id].r <= ri)
	{
		no[id].sum_ += (no[id].r - no[id].l + 1) * delta;/*id的sum_需要保证最新,id子节点的sum_之后可以由id的delta算出*/
		no[id].toadd += delta;
		return;
	}
	if (no[id].toadd)
	{
		int tmp_id = id << 1;
		no[tmp_id].sum_ += (no[tmp_id].r - no[tmp_id].l + 1) * no[id].toadd;
		no[tmp_id].toadd += no[id].toadd;
		++tmp_id;
		no[tmp_id].sum_ += (no[tmp_id].r - no[tmp_id].l + 1) * no[id].toadd;
		no[tmp_id].toadd += no[id].toadd;
		no[id].toadd = 0;
	}
	int tmp_id = id << 1;
	update(le, ri, tmp_id, delta);
	update(le, ri, tmp_id + 1, delta);
	no[id].sum_ = no[tmp_id].sum_ + no[tmp_id + 1].sum_;
	return;
}






int main()
{
	int n = 0, q = 0;
	scanf("%d%d", &n, &q);
	for (int i = 1; i <= n; ++i)
	{
		scanf("%d", &a[i]);
	}
	build(1, n, 1);
	
	char cmd;
	int x = 0, y = 0;
	long long z = 0;
	for (int i = 0; i < q; ++i)
	{
		scanf("\n%[CQ]%d%d", &cmd, &x, &y);
		if (cmd == 'C')
		{
			scanf("%lld", &z);
			update(x, y, 1, z);
		}
		else if (cmd == 'Q')
		{
			ans = 0;
			ask(x, y, 1);
			printf("%lld\n", ans);
		}
	}
	return 0;
}