LightOJ - 1369 （推公式）

LightOJ - 1369 （推公式）

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 nis 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 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

1

3 5

1 2 3

1

0 0 3

1

0 2 1

1

Sample Output

Case 1:

-4

0

4

题目大意：

给出n个数。给出一个f函数。

给出2种操作。

一，0  x   v   把x坐标的值换成v。

二，1输出f函数

思路：

当值发生变化时，推出O（1）计算f函数值的公式。

代码：

#include #include #include #include #include #include #include

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