Element Swapping
Time Limit: 1 Second Memory Limit: 65536 KB
DreamGrid has an integer sequence $a_1,a_2,a_3,\dots ,a_n$ and he likes it very much. Unfortunately, his naughty roommate BaoBao swapped two elements $a_i$ and $a_j$ $(1\le i<j\le n)$ in the sequence when DreamGrid wasn’t at home. When DreamGrid comes back, he finds with dismay that his precious sequence has been changed into $a_1,a_2,\dots ,a_{i-1},a_j,a_{i+1},\dots ,a_{j-1},a_i,a_{j+1},\dots ,a_n$

What’s worse is that DreamGrid cannot remember his precious sequence. What he only remembers are the two values

$x={\sum }_{k=1}^{n}k{a}_k\text{and}y={\sum }_{k=1}^{n}k{a}_k^2$

Given the sequence after swapping and the two values DreamGrid remembers, please help DreamGrid count the number of possible element pairs $(a_i,a_j)$ BaoBao swaps.

Note that as DreamGrid is poor at memorizing numbers, the value of or might not match the sequence, and no possible element pair can be found in this situation.

Two element pairs $(a_i,a_j)(1\le i<j\le n)$ and $(a_p,a_q)(1\le p<q\le n)$ are considered different if $i≠p$ or $j≠q$.

Input
There are multiple test cases. The first line of the input contains an integer $T$, indicating the number of test cases. For each test case:

The first line contains three integers $n$,$x$ and $y$ $(2\le n\le 10^5,1\le x,y\le 10^{18})$, indicating the length of the sequence and the two values DreamGrid remembers.

The second line contains $n$ integers $b_1,b_2,\dots ,b_n(1\le b_i\le 10^5)$, indicating the sequence after swapping. It’s guaranteed that ${\sum }_{k=1}^{n}k{b}_k\le 10^{18}$ and ${\sum }_{k=1}^{n}k{b}_k^2\le 10^{18}$.

It’s guaranteed that the sum of of all test cases will not exceed $2\times 10^6$.

Output
For each test case output one line containing one integer, indicating the number of possible element pairs BaoBao swaps.

Sample Input
2
6 61 237
1 1 4 5 1 4
3 20190429 92409102
1 2 3
Sample Output
2
0
Hint
For the first sample test case, it’s possible that BaoBao swaps the 2nd and the 3rd element, or the 5th and the 6th element.

题目大意：
第一行输入一个整数$t$，代表有$t$组测试样例，对于每组测试样例，先输入三个整数$n,x_1,y_1$，下一行再输入$n$个整数的数组$a_1,a_2,\dots ,a_n$$(1\le a_i\le 10^5)$，现在我们能交换数组中的任意两个元素$a_i$和$a_j$ $(1\le i<j\le n)$，使得改变后的数组求得的$x,y$变为$x1,y1$，问有多少种交换方案。
解题思路：
通过题意我们可以知道，此题相当于我们可以根据公式$x={\sum }_{k=1}^{n}k{a}_k$ 和 公式$y={\sum }_{k=1}^{n}k{a}_k^2$求出当前数组的$x$和$y$，我们可以将其记作$x_2,y_2$，交换数组中的两个元素后的$x,y$为$x_1,y_2$，因为我们仅交换了数组中的两个元素，所以根据$x,y$的公式，我们可以知道$x_1,x_2$的区别在于$x2$中$a_i,a_j$两项为$i{a}_i+j{a}_j$，而$x1$也就是交换后$a_i,a_j$两项为$j{a}_i+i{a}_j$，其他项完全相同，所以可以得到$x2-x1=(i-j)(a_i-a_j)$，同理$y2-y1=(i-j)(a_i^2-a_j^2)$即$y_2-y_1=(i-j)(a_i-a_j)(a_i+a_j)$。
因此可以先假设$x=x_2-x_1,t=y_2-y_1$，如果$x==0\&\&y==0$的话，那么则证明$a_i==a_j$，则方案数则为数组中选取任意两个相同项的方案数，例如假设$a$在数组中出现了$k$次，则$ans+=C_k^2$。
若$x≠0\&\&y≠0$，通过$x=(i-j)(a_i-a_j)$和$y=(i-1)(a_i-a_j)(a_i+a_j)$，可知$y/x=a_i+a_j$，这是就可以判断一下若$y\%x==0$则存在，否则$ans=0$，因此可以得到$a_i+a_j$的值，此时就可以通过遍历数组$a[]$，求出其对应的$a[j]$，通过遍历可以确定$i,a[i],a[j]$的值，然后根据$x=(i-j)(a_i-a_j)$，可以求出$j$，这时只需判断数组中的第$j$项是否为$a[j]$，若是则方案数加一。
代码：

#include <cstdio>
#include <iostream>
#include <algorithm>
#include <cmath>
#include <cstdlib>
#include <cstring>
#include <map>
#include <stack>
#include <queue>
#include <vector>
#include <bitset>
#include <set>
#include <utility>
#include <sstream>
#include <iomanip>
using namespace std;
typedef long long ll;
typedef unsigned long long ull;
#define inf 0x3f3f3f3f
#define rep(i,l,r) for(int i=l;i<=r;i++)
#define lep(i,l,r) for(int i=l;i>=r;i--)
#define ms(arr) memset(arr,0,sizeof(arr))
//priority_queue<int,vector<int> ,greater<int> >q;
const int maxn = (int)1e5 + 5;
const ll mod = 1e9+7;
ll arr[100100];
ll num[100100];
set<ll> s;
int main() 
{#ifndef ONLINE_JUDGEfreopen("in.txt", "r", stdin);#endif//freopen("out.txt", "w", stdout);ios::sync_with_stdio(0),cin.tie(0);int t;scanf("%d",&t);while(t--) {int n;ll x1,x2=0,y1,y2=0;scanf("%d %lld %lld",&n,&x1,&y1);s.clear();memset(num,0,sizeof num);rep(i,1,n) {scanf("%lld",&arr[i]);x2+=(ll)i*arr[i];y2+=(ll)i*arr[i]*arr[i];s.insert(arr[i]);num[arr[i]]++;}ll x=x2-x1,y=y2-y1;ll ans=0;if(x==0&&y==0) {for(auto it=s.begin();it!=s.end();it++) {if(num[*it]>1) {ans+=(num[*it]*(num[*it]-1))/2LL;}}}else if(x!=0&&y!=0) {if(y%x==0) {ll t=y/x;for(int i=1;i<=n;i++) {if(t-arr[i]>0&&t-arr[i]<=100000&&num[t-arr[i]]>0) {ll nape=arr[i]-(t-arr[i]);if(nape!=0&&x%nape==0) {int k=i-x/nape;if(k>i&&k<=n&&arr[k]==t-arr[i]) ans++;}}}}else ans=0;}else ans=0;printf("%lld\n",ans);}return 0;
}