CodeForces 710D Two Arithmetic Progressions(模线性方程组)

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对于给定的两个一元二次方程，问使得L<=a1x1+b1==a2x2+b2<=R且x1,x2>=0的解一共有几个

比较容易看出来对于任意一个解X， X==b1(mod a1),X==b2(mod a2)

那么使用扩展欧几里得得到m0 a0

使得任意k k*m0+a0都是一个可行解

现在问题就变成了对于给定的区间L，R，L<=x*m0+a0<=R x的个数

又因为问题要求 x1,x2>=0 故我们需要处理左边界

#include <cstdio>#include <iostream>#include <algorithm>#include <cstdlib>#include <cmath>#include <string>#include <cstring>#include <string.h>#include <vector>#include <set>#include <map>#include <queue>#include <stack>using namespace std;#define sp system("pause")typedef long long ll;typedef pair<int, int> pii;ll extend_gcd(ll a, ll b, ll &x, ll &y){if (a == 0 && b == 0)return -1;if (b == 0){ x = 1; y = 0; return a; }ll d = extend_gcd(b, a%b, y, x);y -= a / b*x;return d;}ll m[10], a[10];bool solve(ll &m0, ll &a0, ll m, ll a){ll y, x;ll g = extend_gcd(m0, m, x, y);if (abs(a - a0) % g)return false;x *= (a - a0) / g;x %= m / g;a0 = (x*m0 + a0);m0 *= (m / g);a0 %= m0;if (a0 < 0)a0 += m0;return true;}bool MLES(ll &m0, ll &a0, int n){bool flag = true;m0 = 1; a0 = 1;for (int i = 0; i < n; i++){if (!solve(m0, a0, m[i], a[i])){flag = false;break;}}return flag;}int main(){ll L, R; ll b1, b0;cin >> m[0] >> a[0] >> m[1] >> a[1] >> L >> R;b1 = a[1]; b0 = a[0];if (a[1] < 0){ll pl = (-a[1]) / m[1];a[1] = (a[1] + (pl + 1)*m[1]) % m[1];}if (a[0] < 0){ll pl = (-a[0]) / m[0];a[0] = (a[0] + (pl + 1)*m[0]) % m[0];}ll m0, a0;if (!MLES(m0, a0, 2)){puts("0");return 0;}ll gao = max(b1, b0);ll ans = 0;L = max(L, gao);if (L>R){puts("0");return 0;}ll pu = ceil((L - a0)*1.0 / m0);if (L == pu*m0 + a0)ans++, pu++;while (pu*m0 + a0 < L)pu++;L = pu*m0 + a0;if (L <= R){ans++;ans += (R - L) / m0;}printf("%I64d\n", ans);}

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