Round B APAC Test 2017 Problem B. Sherlock and Watson Gym Secrets

Problem B. Sherlock and Watson Gym Secrets

Watson and Sherlock are gym buddies.
Their gym trainer has given them three numbers, A, B, and N, and has asked Watson and Sherlock to pick two different positive integers i and j, where i and j are both less than or equal to N. Watson is expected to eat exactly i^A sprouts every day, and Sherlock is expected to eat exactly j^B sprouts every day.
Watson and Sherlock have noticed that if the total number of sprouts eaten by them on a given day is divisible by a certain integer K, then they get along well that day.
So, Watson and Sherlock need your help to determine how many such pairs of (i, j) exist, where i != j. As the number of pairs can be really high, please output it modulo 10⁹[phone]).

Input

The first line of the input gives the number of test cases, T. T test cases follow. Each test case consists of one line with 4 integers A, B, N and K, as described above.

Output

For each test case, output one line containing Case #x: y, where x is the test case number (starting from 1) and y is the required answer.

Limits

1 ≤ T ≤ 100.
0 ≤ A ≤ 10⁶.
0 ≤ B ≤ 10⁶.

Small dataset

1 ≤ K ≤ 10000.
1 ≤ N ≤ 1000.

Large dataset

1 ≤ K ≤ 100000.
1 ≤ N ≤ 10¹⁸.

Sample

Input 
 
Output 
 

31 1 5 31 2 4 51 1 2 2

Case #1: 8Case #2: 3Case #3: 0

In Case 1, the possible pairs are (1, 2), (1, 5), (2, 1), (2, 4), (4, 2), (4, 5), (5, 1), and (5, 4).
In Case 2, the possible pairs are (1, 2), (1, 3), and (4, 1).
In Case 3, No possible pairs are there, as i != j.

Solution

#include <iostream>#include <cstdio>#include <cstring>#include <string>#include <vector>#include <stack>#include <list>#include <iterator>#include <unordered_set>#include <unordered_map>using namespace std;typedef long long ll;const int MAXK = 1e5;ll i_A[MAXK+1], j_B[MAXK+1], i_j_same[MAXK+1];int pow_mod(int a, int n, int mod){    int r = 1 % mod;    while (n) {        if (n&1)            r = (ll)r*a%mod;        a = (ll)a*a%mod;        n >>= 1;    }    return r;}int main(int argc, char* argv[]){#ifndef TEST    if (argc != 2)    {        cout << "Invalid input" << endl;        return 1;    }    string input = argv[1];    string output = input.substr(0, input.length() - 2) + "out";    freopen(input.c_str(), "r", stdin);    freopen(output.c_str(), "w", stdout);#endif    const ll MOD = 1e9 + 7;    int T;    int A, B;    ll N, K;    cin >> T;    for (int i = 1; i <= T; i++)    {        memset(i_A, 0, sizeof(i_A));        memset(j_B, 0, sizeof(j_B));        memset(i_j_same, 0, sizeof(i_j_same));        cin >> A >> B >> N >> K;        ll res = 0;        for (int p = 1; p <= min(N, K); p++)        {            ll cnt = ((N - p) / K + 1) % MOD;            int tmp_A = pow_mod(p, A, K), tmp_B = pow_mod(p, B, K);            i_A[tmp_A] += cnt;            j_B[tmp_B] += cnt;            if ((tmp_A + tmp_B) % K == 0) // (0 + 0) != K, but (0 + 0) % K == 0                i_j_same[p%K] += cnt;        }        for (int p = 0; p < K; p++)        {            int q = (K - p) % K; // Convert (0, K) to (0, 0)            res += ((i_A[p] % MOD) * (j_B[q] % MOD)) % MOD;            res -= i_j_same[p];            res %= MOD;        }        res += MOD;        res %= MOD;        cout << "Case #" << i << ": " << res << endl;    }    fclose(stdin);    fclose(stdout);    return 0;}

Note

首先要会计算Fast Modular Exponentiation，见我另一篇博客http://blog.csdn.net/ywcpig/article/details/52566496。

其次，要知道数论中同余的基本性质（同加、同乘）。

本题答案 = 所有(i^A+j^B)%K==0的情况数量 - i==j的特例数量

注意到：((i+K)^A)%K == (i^A)%K
所以，只需要考虑i=1到K的情况就行。

综上，本题解法的时间复杂度为O(KlogK)

Reference

https://code.google.com/codejam/contest/5254487/dashboard#s=p1