The daily life about a frog.
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
[CODE_BLOCK_0]| c++
#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;}
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