lightoj Answering Queries 1369 （数学转换&&技巧）

Answering Queries

Time Limit: 3000MSMemory Limit: 32768KB64bit IO Format: %lld & %llu

Submit Status

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 10⁶denoting 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

//花了两个小时搞这个题，虽然浪费不少时间，但做出来了，还是有一点小小的喜悦感的。。。

 

#include<stdio.h>#include<string.h>long long a[100010];int main(){int t;int T=1;int n,m,c,x,y;long long sum;int i,j,k;scanf("%d",&t);while(t--){scanf("%d%d",&n,&m);for(i=0;i<n;i++)scanf("%lld",&a[i]);j=0;sum=0;for(i=n-1;j<n;i-=2)sum+=i*a[j++];printf("Case %d:\n",T++);while(m--){scanf("%d",&c);if(c)printf("%lld\n",sum);else{scanf("%d%d",&x,&y);sum+=(n-1-2*x)*(y-a[x]);a[x]=y;}}}return 0;}