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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#632115 | #7735. Primitive Root | Lynia | WA | 0ms | 3712kb | C++23 | 2.8kb | 2024-10-12 12:15:19 | 2024-10-12 12:15:20 |
Judging History
answer
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#pragma GCC optimize(2)
#include <iostream>
#include <iomanip>
#include <algorithm>
#include <map>
#include <set>
#include <queue>
#include <string>
#include <cstring>
#include <cmath>
#include <list>
#include <stack>
#include <array>
#include <unordered_map>
#include <unordered_set>
#include <bitset>
#include <random>
#include <numeric>
#include <functional>
//#include <Windows.h>
using namespace std;
#define fa(i,op,n) for (int i = op; i <= n; i++)
#define fb(j,op,n) for (int j = op; j >= n; j--)
#define pb push_back
#define HashMap unordered_map
#define HashSet unordered_set
#define var auto
#define all(i) i.begin(), i.end()
#define endl '\n'
#define px first
#define py second
#define DEBUG cout<<-1<<endl
using VI = vector<int>;
using VL = vector<long long>;
using ll = long long;
using ull = unsigned long long;
using db = double;
using pii = pair<int, int>;
using pll = pair<ll, ll>;
template<class T1, class T2>
ostream& operator<<(ostream& out, const pair<T1, T2>& p) {
out << '(' << p.first << ", " << p.second << ')';
return out;
}
template<typename T>
ostream& operator<<(ostream& out, const vector<T>& ve) {
for (T i : ve)
out << i << ' ';
return out;
}
template<class T1, class T2>
ostream& operator<<(ostream& out, const map<T1, T2>& mp) {
for (auto i : mp)
out << i << '\n';
return out;
}
const int INF = 0x3f3f3f3f;
const ll LNF = 0x3f3f3f3f3f3f3f3f;
const int IINF = 0x7fffffff;
const int iinf = 0x80000000;
const ll LINF = 0x7FFFFFFFFFFFFFFF;
const ll linf = 0x8000000000000000;
int dx[8] = { 1, -1, 0, 0, 1, -1, 1, -1 };
int dy[8] = { 0, 0, 1, -1, 1, -1, -1, 1 };
//#include "Lynia.h"
const int mod = 998244353;
const int N = 5e3 + 10;
void solve(int CASE)
{
ll p, m; cin >> p >> m;
ll ans = m / p;
// p(k - 1) + 2 <= g <= p(k + 1),g <= m
// k 都合法 p(k + 1) <= m:
// k <= m / p - 1
// k 都不合法 p(k - 1) + 2 > m:
// k > ceil((m - 2) / p) + 1
// 枚举中间值
ll l = m / p - 1 + 1, r = (m - 2 + p - 1) / p + 1;
fa(i, l, r) {
ans += ((i * 1ll ^ (p - 1)) % p == 1);
}
cout << ans << endl;
return;
}
int main()
{
//SetConsoleOutputCP(CP_UTF8);
ios::sync_with_stdio(0), cin.tie(0), cout.tie(0);
int _ = 1;
cin >> _;
fa(i, 1, _)solve(i);
return 0;
}
Details
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Test #1:
score: 0
Wrong Answer
time: 0ms
memory: 3712kb
input:
3 2 0 7 11 1145141 998244353
output:
1 1 871
result:
wrong answer 2nd lines differ - expected: '2', found: '1'