题目链接：
http://lightoj.com/volume_showproblem.php?problem=1369
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.
思路：把这N个数写到一个N*N的矩阵的对角线上，然后后面的减出来的数就写在后面，类似于下面的图，这样就可以发现，如果你改变某个数的值，例如咱们增加了他的值A，那么影响就是：这一行的每个数都+A，这一列的数都-A，这样就可以了，我们只要统计增加的个数就可以了，不要忘了更新数组
公式：
对于每个a[i]
sum+=(n-i-1)*a[i]-i*a[i];
n-i-1表示行，i表示列
This is the code：
#include<algorithm>
#include<cstdio>
#include<cstring>
#include<cmath>
#include<cstdlib>
#include<iostream>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<sstream>
#include<stack>
#include<string>
#include<set>
#include<vector>
using namespace std;
#define PI acos(-1.0)
#define EPS 1e-8
#define LL long long
#define ULL unsigned long long //1844674407370955161
#define INT_INF 0x7f7f7f7f //2139062143
#define LL_INF 0x7f7f7f7f7f7f7f7f //9187201950435737471
// ios::sync_with_stdio(false);
// 那么cin, 就不能跟C的 scanf，sscanf, getchar, fgets之类的一起使用了。
const int dr[]= {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[]= {-1, 1, 0, 0, -1, 1, -1, 1};
int a[1000050];
int main()
{int T;scanf("%d",&T);for(int t=1; t<=T; ++t){LL sum=0;LL n,k;scanf("%lld%lld",&n,&k);for(LL i=0; i<n; i++){scanf("%lld",&a[i]);sum+=(n-i-1ll)*a[i]-i*a[i];//计算初始的和//n-i-1表示行，i表示列}printf("Case %d:\n",t);int tem;while(k--){scanf("%d",&tem);if(!tem){LL x,v;scanf("%lld%lld",&x,&v);LL y=v-a[x];sum+=((n-x-1ll)*y-x*y);a[x]=v;//不要忘了更新}elseprintf("%lld\n",sum);}}return 0;
}