Gửi bài giải
Điểm:
2,00 (OI)
Giới hạn thời gian:
2.25s
Giới hạn bộ nhớ:
512M
Input:
geometric.inp
Output:
geometric.out
Tác giả:
Nguồn bài:
Dạng bài
Ngôn ngữ cho phép
C, C++, Go, Java, Kotlin, Pascal, PyPy, Python, Rust, Scratch
Input
Output
Sample Input 1
2
13 15
Sample Output 1
7 8
Bình luận
include<bits/stdc++.h>
define FOR(i, a, b) for (int i = (a), _b = (b); i <= _b; i++)
define FORD(i, b, a) for (int i = (b), _a = (a); i >= _a; i--)
define REP(i, n) for (int i = 0, _n = (n); i < _n; i++)
define FORE(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); i++)
define ALL(v) (v).begin(), (v).end()
define fi first
define se second
define MASK(i) (1LL << (i))
define BIT(x, i) (((x) >> (i)) & 1)
define div _div
define next _next
define prev _prev
define left _left
define right _right
define _builtinpopcount _builtinpopcountll
using namespace std; template<class X, class Y> bool minimize(X &x, const Y &y) { X eps = 1e-9; if (x > y + eps) { x = y; return true; } else return false; } template<class X, class Y> bool maximize(X &x, const Y &y) { X eps = 1e-9; if (x + eps < y) { x = y; return true; } else return false; } template<class T> T Abs(const T &x) { return (x < 0 ? -x : x); }
/* Author: Van Hanh Pham */
/* END OF TEMPLATE - ACTUAL SOLUTION COMES HERE */
define MAX 20002000
define SQRT 4545
define LENGTH 25
const int MOD = (int)1e9 + 22071997;
int result[MAX + 3], pw[SQRT + 3][LENGTH + 3]; bool coprime[SQRT + 3][SQRT + 3]; int primeDiv[MAX + 3]; long long savedResult[MAX + 3];
int getPw(int x, int k) { if (k == 0) return 1; if (k == 1) return x; return x > SQRT ? MAX + 1 : pw[x][k]; }
int getSum(int p, int q, int k) { long long res = 0; REP(i, k + 1) { res += 1LL * getPw(p, i) * getPw(q, k - i); if (res > MAX) return MAX; } return res; }
void prepare(void) { REP(i, 2) primeDiv[i] = -1; for (int i = 2; i * i <= MAX; i++) if (primeDiv[i] == 0) for (int j = i * i; j <= MAX; j += i) primeDiv[j] = i; FOR(i, 2, MAX) if (primeDiv[i] == 0) primeDiv[i] = i;
}
vector<pair> factors;
void backtrack(int pos, int val, long long &sum) {
if ((int)pos >= factors.size()) {
sum += result[val];
return;
}
}
int solve(int n) { if (n < 3) return 0;
}
int main(void) {
ifdef ONLINE_JUDGE
endif // ONLINE_JUDGE
}
/* LOOK AT MY CODE. MY CODE IS AMAZING :D */
code nha mọi người
ai bt ko chỉ tui vs
include<bits/stdc++.h>
define FOR(i, a, b) for (int i = (a), _b = (b); i <= _b; i++)
define FORD(i, b, a) for (int i = (b), _a = (a); i >= _a; i--)
define REP(i, n) for (int i = 0, _n = (n); i < _n; i++)
define FORE(i, v) for (__typeof((v).begin()) i = (v).begin(); i != (v).end(); i++)
define ALL(v) (v).begin(), (v).end()
define fi first
define se second
define MASK(i) (1LL << (i))
define BIT(x, i) (((x) >> (i)) & 1)
define div _div
define next _next
define prev _prev
define left _left
define right _right
define _builtinpopcount _builtinpopcountll
using namespace std; template<class X, class Y> bool minimize(X &x, const Y &y) { X eps = 1e-9; if (x > y + eps) { x = y; return true; } else return false; } template<class X, class Y> bool maximize(X &x, const Y &y) { X eps = 1e-9; if (x + eps < y) { x = y; return true; } else return false; } template<class T> T Abs(const T &x) { return (x < 0 ? -x : x); }
/* Author: Van Hanh Pham */
/* END OF TEMPLATE - ACTUAL SOLUTION COMES HERE */
define MAX 20002000
define SQRT 4545
define LENGTH 25
const int MOD = (int)1e9 + 22071997;
int result[MAX + 3], pw[SQRT + 3][LENGTH + 3]; bool coprime[SQRT + 3][SQRT + 3]; int primeDiv[MAX + 3]; long long savedResult[MAX + 3];
int getPw(int x, int k) { if (k == 0) return 1; if (k == 1) return x; return x > SQRT ? MAX + 1 : pw[x][k]; }
int getSum(int p, int q, int k) { long long res = 0; REP(i, k + 1) { res += 1LL * getPw(p, i) * getPw(q, k - i); if (res > MAX) return MAX; } return res; }
void prepare(void) { REP(i, 2) primeDiv[i] = -1; for (int i = 2; i * i <= MAX; i++) if (primeDiv[i] == 0) for (int j = i * i; j <= MAX; j += i) primeDiv[j] = i; FOR(i, 2, MAX) if (primeDiv[i] == 0) primeDiv[i] = i;
}
vector<pair> factors;
void backtrack(int pos, int val, long long &sum) {
if ((int)pos >= factors.size()) {
sum += result[val];
return;
}
}
int solve(int n) { if (n < 3) return 0;
}
int main(void) {
ifdef ONLINE_JUDGE
endif // ONLINE_JUDGE
}
/* LOOK AT MY CODE. MY CODE IS AMAZING :D */
...
Các bạn xem đề bài ở đây nhé: https://oj.vnoi.info/problem/pvh5_a
đề mình tìm ở đâu v mn