Light OJ 1369 Answering Queries （思维）-优快云博客

本文介绍了一个关于快速处理数组中特定查询的问题解决方案。通过对题目给定函数的深入分析，找到了一种优化的方法来避免超时问题。文章详细解释了如何通过观察规律来减少计算量，并提供了一段示例代码。

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H - Answering Queries

Time Limit:3000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu

Submit Status Practice LightOJ 1369

uDebug

Appoint description:

Description

The problem you need to solve here is pretty simple. You are give a function f(A, n), where A is an array of integers and n is the number of elements in the array. f(A, n) is defined as follows:

long long f( int A[], int n ) { // n = size of A

long long sum = 0;

for( int i = 0; i < n; i++ )

for( int j = i + 1; j < n; j++ )

sum += A[i] - A[j];

return sum;

}

Given the array A and an integer n, and some queries of the form:

1) 0 x v (0 ≤ x < n, 0 ≤ v ≤ 10⁶), meaning that you have to change the value of A[x] to v.

2) 1, meaning that you have to find f as described above.

Input

Input starts with an integer T (≤ 5), denoting the number of test cases.

Each case starts with a line containing two integers: n and q (1 ≤ n, q ≤ 10⁵). The next line contains n space separated integers between 0 and 10⁶denoting the array A as described above.

Each of the next q lines contains one query as described above.

Output

For each case, print the case number in a single line first. Then for each query-type "1" print one single line containing the value of f(A, n).

Sample Input

3 5

1 2 3

0 0 3

0 2 1

Sample Output

Case 1:

-4

Hint

Dataset is huge, use faster I/O methods.

Problem Setter: Hasnain Heickal Jami

Special Thanks: Jane Alam Jan

Developed and Maintained by
JANE ALAM JAN

思路：很明显不能按照题目给的函数写，会超时，那么就只能找规律了，当i=0时，有a0-a1+a0-a2+a0-a3+.....，当i=1时，有a1-a2+a1-a3+a1+.....,观察可得发现第一个式子刚好比第二个多出来 n-i 个（a0-a1)。之后操作时要改变值时直接在之前求得的总和身上改，例如改变了 a7 增加了2 则从第8到n 个数被 a7减时结果都要多2，而从第0个到第6个减 a7 时结果都小2 。

#include<stdio.h>
#include<string.h>
#include<algorithm>
#define LL long long
using namespace std;

LL a[101100];
LL init(LL a[],LL n)
{
	LL sum=0,temp=0;
	LL i;
	for(i=1;i<n;i++)
	temp+=a[0]-a[i];
	sum=temp;
	for(i=1;i<n;i++)
	{
		LL cnt=a[i]-a[i-1];
		temp+=cnt*(n-i);
		sum+=temp;
	 } 
	 return sum;
 }
int main()
{
 	LL t,n,q,m,b,c,i;
 	scanf("%lld",&t);
 	LL k=1;
 	while(t--)
 	{
 		scanf("%lld%lld",&n,&q);
 		for(i=0;i<n;i++)
 		scanf("%lld",&a[i]);
 	    LL sum=init(a,n);
 		printf("Case %lld:\n",k++);
		while(q--)
		{
			scanf("%lld",&m);
			if(m==1)
			printf("%lld\n",sum);
			else
			{
				scanf("%lld%lld",&b,&c);
				LL temp=c-a[b];
				a[b]=c;
				sum+=temp*(n-b-b-1);
			}
		} 
	 }
}