Time Limit: 3000MS

Memory Limit: 32768KB

64bit IO Format: %lld & %llu

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 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

Source

Problem Setter: Hasnain Heickal Jami

Special Thanks: Jane Alam Jan

 

细节+ 规律,

#include 
       
        #define N 100000+100typedef long long LL;LL a[N];int main(){    int t, Q=1; scanf("%d", &t);    while(t--)    {        int n, q; scanf("%d%d", &n, &q);        for(int i=0; i
        
          n)                continue;            getSum +=A*a[i]-b*a[i];            A--; b++;         }        int firNum;        printf("Case %d:\n", Q++);        while(q--)        {            scanf("%d", &firNum);            if(firNum==1)                printf("%lld\n", getSum);            else            {                int posi; LL modify;                scanf("%d%lld", &posi, &modify);                LL wc=modify-a[posi];                getSum=getSum+(n-1-2*posi)*wc;                a[posi]=modify;  //更新必须要有 !!!!! ;            }        }    }    return 0;}