light oj 1369 - Answering Queries

                                                                                                                        1369 - Answering Queries

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Time Limit: 3 second(s)Memory Limit: 32 MB

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

Output for Sample Input

1

3 5

1 2 3

1

0 0 3

1

0 2 1

1

Case 1:

-4

0

4

Note

Dataset is huge, use faster I/O methods.

 #include<cstdio> #include<iostream> using namespace std; long long a[100010]; int main() {     int t,n,m,l = 1;     int k,x,v;     scanf("%d",&t);     while(t--)     {         scanf("%d%d",&n,&m);         long long sum = 0;         for(int i = 0 ; i < n ; i++)         {             scanf("%lld",&a[i]);             sum += a[i];         }         long long ans = 0;         for(int i = 0 ; i < n ; i++)        {            ans += (a[i] * (n-i-1) - (sum-a[i]));  // 计算f（A[]，n） 的值               sum -= a[i];            }         printf("Case %d:\n",l++);         while(m--)         {             scanf("%d",&k);             if(k == 1)                printf("%lld\n",ans);             else             {                 scanf("%d%d",&x,&v);                 ans -= ((v-a[x]) * x + (a[x]-v) * (n-x-1)); // 考虑将A[x]换成V对ans和的影响                 a[x] = v;   //更新改变后的值             }         }     }     return 0; }