HDU 6128 Inverse of sum(推公式)

最新推荐文章于 2020-02-04 23:31:45 发布

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最新推荐文章于 2020-02-04 23:31:45 发布

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本文介绍了一种算法，用于解决特定的数学问题：给定一个整数序列和一个质数p，找出序列中所有满足特定模逆数性质的数对。通过巧妙地变换等式并使用哈希表优化搜索过程，该算法能够在合理的时间内解决问题。

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Inverse of sum

Time Limit: 6000/3000 MS (Java/Others) Memory Limit: 524288/524288 K (Java/Others)
Total Submission(s): 657 Accepted Submission(s): 223

Problem Description
There are n nonnegative integers a1…n which are less than p. HazelFan wants to know how many pairs i,j(1≤i<j≤n) are there, satisfying 1ai+aj≡1ai+1aj when we calculate module p, which means the inverse element of their sum equals the sum of their inverse elements. Notice that zero element has no inverse element.

Input
The first line contains a positive integer T(1≤T≤5), denoting the number of test cases.
For each test case:
The first line contains two positive integers n,p(1≤n≤105,2≤p≤1018), and it is guaranteed that p is a prime number.
The second line contains n nonnegative integers a1…n(0≤ai<p).

Output
For each test case:
A single line contains a nonnegative integer, denoting the answer.

Sample Input
2
5 7
1 2 3 4 5
6 7
1 2 3 4 5 6

Sample Output
4
6

题意

一个长度为n的序列，问其中有几对 $1ai+aj≡1ai+1aj\frac{1}{a_{i}+a_{j}}\equiv \frac{1}{a_{i}}+\frac{1}{a_{j}}$ (mod p)

思路

$1ai+aj≡1ai+1aj\frac{1}{a_{i}+a_{j}}\equiv \frac{1}{a_{i}}+\frac{1}{a_{j}}$ *(mod p)两边同乘 $a_{i}+a_{j}$ 得
$\equiv 1+\frac{a_{j}}{a_{i}}+1+\frac{a_{i}}{a_{j}}$ (mod p)再同乘 $a_{i}a_{j}$ 得
$ai2+aiaj+aj2≡0a_{i}^{2}+a_{i}a_{j}+a_{j}^{2}\equiv 0$ (mod p)*再同乘 $a_{i}-a_{j}$

$ai3−aj3≡0a_{i}^{3}-a_{j}^{3}\equiv 0$ (mod p)
然后我们枚举 $a_{i}^{3}$ %p用map来储存
值得注意的是当 $a_{i}==a_{j}$ 时 $ai2+aiaj+aj2≡0a_{i}^{2}+a_{i}a_{j}+a_{j}^{2}\equiv 0$ 就变为 $3ai2≡03a_{i}^{2}\equiv 0$ ，而此时若 $ai2+aiaj+aj2≡0a_{i}^{2}+a_{i}a_{j}+a_{j}^{2}\equiv 0$ 不成立而还是乘上了一个 $a_{i}-a_{j}$ ，那么我们把这个时候的 $a_{i}^{3}$ %p算出来也无用，所以要去掉当 $ai2+aiaj+aj2≡0a_{i}^{2}+a_{i}a_{j}+a_{j}^{2}\equiv 0$ 不成立的时候的 $a_{i}$ 即 $3ai2≡03a_{i}^{2}\equiv 0$ 不成立的情况

#include <stdio.h>
#include <string.h>
#include <iostream>
#include <algorithm>
#include <map>
using namespace std;
map<long long,int>num;
map<long long,int>cnt;
long long multi(long long a,long long b,long long p){
    long long ans=0;
    while(b){
        if(b%2==1)
            ans=(ans+a)%p;
        b/=2;
        a=(a+a)%p;
    }
    return ans;
}
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        long long n,p;
        scanf("%lld%lld",&n,&p);
        cnt.clear();
        num.clear();
        long long ans=0;
        for(int i=0;i<n;i++)
        {
            long long a;
            scanf("%lld",&a);
            if(a==0)
                continue;
            if(multi(multi(a,a,p),3,p)!=0)
                ans-=cnt[a];
            cnt[a]++;
            long long val=multi(multi(a,a,p),a,p);
            ans+=num[val]++;
        }
        printf("%lld\n",ans);
    }
    return 0;
}