lightoj1369

maksyuki 发表于 oj 分类，标签:
27 8月 2015
0

1369 - Answering Queries

long long f( int A[], int n ) { // n = size of AThe 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 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 106 denoting 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

Output for Sample Input
13 51 2 310 0 310 2 1

1

 Case 1:-404

Note

Dataset is huge, use faster I/O methods.

 

题目类型：简单数学

算法分析：首先将sum求出来，然后更新时只更新a[x]的值所引起的变化，查询时直接输出结果，注意两个int型变量相乘的中间结果也是int的，可能会溢出！！！