poj3468 A Simple Problem with Integers

A Simple Problem with Integers

Time Limit: 5000MS		Memory Limit: 131072K
Total Submissions: 80166		Accepted: 24744
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

苟蒻上周才学的这题忘记写博客瞬间忘了什么鬼？！！！！

好吧。。其实线段树轻松AC但是学长执意要我学树状数组

树状数组不是只支持单点修改吗(hehehe)

设sum[i]表示前i项前缀和，c1[i]代表从i项开始到n的变化量,x为查询序列长度

于是ans = sum[r]-sum[l-1]+c1[l]*x+c1[l+1]*(x-1)+c1[l+2]*(x-2)+...+c1[r]*1

          = sum[r]-sum[l-1]+segema(c1[i]*(r-i+1))

      = sum[r]-sum[l-1]+(r+1)*segema(c1[i])-i*segema(c1[i])

那再设个c2[i]=i*c1[i]再加个树状数组就完成啦！

#include<cstdio>
#include<iostream>
#include<algorithm>
#include<cstring>
#include<string>
#include<queue>
#include<stack>
#include<vector>
#include<cstdlib>
#include<map>
#include<cmath>
#include<set>
using namespace std;

const int maxn = 1e5 + 10;
typedef long long LL;

LL c1[maxn],c2[maxn],sum[maxn];
int i,j,n,m;

LL Sum(int r)
{
	LL ret = 0;
	int y;
	LL x = r;
	for (y = r; y > 0; y -= y&-y)
	  ret += ((x+1)*c1[y] - c2[y]);
	 return ret; 
}

void Modify (int y,LL k)
{
	LL d = y;
	for (; y <= n; y += y&-y)
	  c1[y] += k,c2[y] += d*k; 
}

int main()
{
	#ifndef ONLINE_JUDGE
	#ifndef YZY
	  freopen(".in","r",stdin);
	  freopen(".out","w",stdout);
	#else
	  freopen("yzy.txt","r",stdin);
	#endif
	#endif
	cin >> n >> m;
	for (i = 1; i <= n; i++) scanf("%lld",&sum[i]),sum[i] += sum[i-1];
	while (m--)
	{
		int l,r;
		LL k;
		char c[2];
		scanf("%s",c);
		if (c[0] == 'Q')
		{
			scanf("%d%d",&l,&r);
			LL ans = sum[r] - sum[l-1];
			ans += Sum(r) - Sum(l-1);
			printf("%lld\n",ans);
		}
		else 
		{
			scanf("%d%d%lld",&l,&r,&k);
			Modify(l,k);
			Modify(r+1,-k);
		}
	} 
	return 0;
}