lightoj 1369 Answering Queries

                                                   Answering Queries

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

根据性质，推公式   

当a[x] 改成v 时    ans 增加了  （n-1-x）*（v-a[x]）+  x *(a[x] - x)  

<pre name="code" class="cpp">#include <iostream>      #include <cstdio> #include <cstring>using namespace std;const int maxn=100005;long long a[maxn],sum,ans,v;int n,q,op,x;int main(){    int T;    scanf("%d",&T);    for(int co=1;co<=T;co++)    {         scanf("%d %d",&n,&q);         for(int i=0;i<n;i++)  scanf("%lld",&a[i]);         sum=a[n-1],ans=0;         for(int i=n-2,j=1;i>=0;i--,j++)         {              ans+=j*a[i]-sum;              sum+=a[i];         }         printf("Case %d:\n",co);         for(int i=0;i<q;i++)         {              scanf("%d",&op);              if(op==1)  printf("%lld\n",ans);              else              {                   scanf("%d %lld",&x,&v);                   ans=ans+(n-1-x)*(v-a[x])+x*(a[x]-v);                   a[x]=v;              }         }    }    return 0;}