练习codeforces 1541B. Pleasant Pairs

题目如下
B. Pleasant Pairs
time limit per test2 seconds
memory limit per test256 megabytes
You are given an array 𝑎1,𝑎2,…,𝑎𝑛 consisting of 𝑛 distinct integers. Count the number of pairs of indices (𝑖,𝑗) such that 𝑖<𝑗 and 𝑎𝑖⋅𝑎𝑗=𝑖+𝑗.

Input
The first line contains one integer 𝑡 (1≤𝑡≤104) — the number of test cases. Then 𝑡 cases follow.

The first line of each test case contains one integer 𝑛 (2≤𝑛≤105) — the length of array 𝑎.

The second line of each test case contains 𝑛 space separated integers 𝑎1,𝑎2,…,𝑎𝑛 (1≤𝑎𝑖≤2⋅𝑛) — the array 𝑎. It is guaranteed that all elements are distinct.

It is guaranteed that the sum of 𝑛 over all test cases does not exceed 2⋅105.

Output
For each test case, output the number of pairs of indices (𝑖,𝑗) such that 𝑖<𝑗 and 𝑎𝑖⋅𝑎𝑗=𝑖+𝑗.

题目大意
定义目标指数满足𝑎𝑖⋅𝑎𝑗=𝑖+𝑗.，求出给定数组中这样的指数 (𝑖,𝑗) 的数量

解题思路
考虑到时间复杂度等，用朴素的暴力解法确实行不通，所以为了顺利ac，我们考虑在比较查找时减少次数
原型是a[i] * a[j] == i + j，
把它变形得到a[i] * a[j] - i == j，
我们又能得出a[i] * a[j] == i + j => i + j ≡ 0 (mod a[i]) => j = -I ( mod a[I] ) => j ≡ a[i] - i (mod a[I]) ;
所以在二重循环里的 j 的表达式可以是 j ≡ a[i] - i
同时 j 的 步长也可以变为 j += a[i]；
这样就大大降低了枚举的个数与速度，

点击查看代码

for(int i = 1; i <= n; ++i){
            for (long long j = a[i] - i; j <= n; j += a[i]) {
                if (j <= i)
                    continue;
                if (a[i] * a[j] == i + j)
                    count++;
            }
        }

完整代码

点击查看代码

#include <stdio.h>

int main() {
    int t;
    scanf("%d", &t);
    while (t--){
        int n;
        scanf("%d", &n);
        long long a[200005];
        for(int i = 1; i <= n; ++i){
            scanf("%lld", &a[i]);
        }
        long long count = 0;
        for(int i = 1; i <= n; ++i){
            for (long long j = a[i] - i; j <= n; j += a[i]){
                if (j <= i)
                    continue;
                if (a[i] * a[j] == i + j)
                    count++;
            }
        }
        printf("%lld\n",count);
    }
    return 0;
}

算法参考https://www.cnblogs.com/RioTian/p/14983249.html 简直是天才