PKU 1061 - 青蛙的约会 (扩展欧几里得)
思路
了解了一下扩展欧几里得。。
设经过的时间为t。
$ (X + mt) - (Y + nt) = kL$，化简得$(n - m)t + Ly = X - Y$，符合$ax+by=c$的形式，可以用扩展欧几里得解出$ax+by=gcd(a, b)$的时候的一个解$x_0$。
两边同时乘以$\frac{c}{d}$，可得一个解x。
但是这时候的x可能是负数，因为通解为$x + k* \frac{b} {g}$，所以对$\frac{b}{g}$取余就行
代码
#include <stack>
#include <cstdio>
#include <list>
#include <set>
#include <iostream>
#include <string>
#include <vector>
#include <queue>
#include <functional>
#include <cstring>
#include <algorithm>
#include <cctype>
#include <string>
#include <map>
#include <cmath>
using namespace std;
#define LL long long
#define ULL unsigned long long
#define SZ(x) (int)x.size()
#define Lowbit(x) ((x) & (-x))
#define MP(a, b) make_pair(a, b)
#define MS(arr, num) memset(arr, num, sizeof(arr))
#define PB push_back
#define X first
#define Y second
#define ROP freopen("input.txt", "r", stdin);
#define MID(a, b) (a + ((b - a) >> 1))
#define LC rt << 1, l, mid
#define RC rt << 1|1, mid + 1, r
#define LRT rt << 1#define RRT rt << 1|1
#define BitCount(x) __builtin_popcount(x)
#define BitCountll(x) __builtin_popcountll(x)
#define LeftPos(x) 32 - __builtin_clz(x) - 1
#define LeftPosll(x) 64 - __builtin_clzll(x) - 1const double PI = acos(-1.0);
const int INF = 0x3f3f3f3f;
const double eps = 1e-8;
const int MAXN = 4e5 + 10;const int MOD = 1000007;
const int dir[][2] = { {1, 0}, {0, 1} };int cases = 0;typedef pair<int, int> pii;
typedef vector<int>::iterator viti;
typedef vector<pii>::iterator vitii;
void gcd(LL a, LL b, LL &g, LL &x, LL &y){
if (!b) { g = a; x = 1; y = 0; } else { gcd(b, a%b, g, y, x); y -= x * (a/b); }} int main(){ LL X, Y, m, n, L; while (cin >> X >> Y >> m >> n >> L) { LL g, x, y; LL b = L; gcd(n-m, L, g, x, y); cout << x << endl; cout << y << endl; if ((X-Y) % g != 0) cout << "Impossible" << endl; else { LL p = b / g; x = (X-Y) / g * x; cout << (x % p + p) % p << endl; } } return 0;}