ZOJ

题目地址

题目大意： 有一个n个数的序列，交换其中两个Ai与Aj，给你交换前后的两种关系。 问有多少种情况。

题解 ： https://blog.csdn.net/u011815404/article/details/89607866   （写的很好！  懒得写公式了。。。.）

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 2e5+5;

int T, n;
ll a[maxn];
ll x, y, sum;
map<int, int> cnt;

int main()
{
    scanf("%d",&T);
    while(T--) {
        cnt.clear();  
        sum = 0;
        scanf("%d%lld%lld",&n, &x, &y);
        for(int i = 1; i <= n; i++) {
            scanf("%lld",&a[i]);
            cnt[a[i]]++;
            x -= 1ll * i * a[i];
            y -= 1ll * i * a[i] * a[i];
        }
        if(x == 0) {
            if(y != 0) {
                printf("0
");
                continue;
            }
            for(int i = 1; i <= n; i++)    
                sum += cnt[a[i]]-1; 
        } else {
            if(y % x != 0) {
                printf("0
");
                continue;
            } else {
                ll num = y / x;
                for(int i = 1; i <= n; i++) {
                    ll Aj = num - a[i];
                //    cout << "Aj = " << Aj << endl;
                    if(Aj == a[i]) continue;
                    ll j = (x-Aj*i+a[i]*i)/(a[i]-Aj);
                //    cout << "j = " << j << endl;
                    if(j < 0) continue;
                    if(a[j] == Aj) sum++;
                } 
            }
        }
        printf("%lld
",sum/2);
    }
    return 0;
}