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

Answering Queries

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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;
}