题意：求R-L区间满足x=a1*k+b1=a2*l+b2的x的个数;
思路：求出最小的满足条件的x0，则ans=(L-x)/(a1/gcd(a1, a2)*a2)+1;
注意剪枝，不然会超时；
代码：
1 #include <bits/stdc++.h> 2 #define ll long long 3 #define MAXN 100000+10 4 using namespace std; 5 6 ll get_gcd(ll a, ll b) 7 { 8 return b ? get_gcd(b, a%b) : a; 9 } 10 11 int main(void) 12 { 13 ll a1, b1, a2, b2, R, L; 14 cin >> a1 >> b1 >> a2 >> b2 >> R >> L; 15 ll k=ceil((R-b1)*1.0/a1), l=ceil((R-b2)*1.0/a2); 16 k=max(0ll, k), l=max(0ll, l); 17 ll x1=a1*k+b1, x2=a2*l+b2; 18 // cout << x1 << " " << x2 << endl;//******* 19 if((b2-b1)%get_gcd(a1, a2)!=0) 20 { 21 cout << "0" << endl; 22 return 0; 23 } 24 if(a1==a2&&abs(x1-x2)<a1&&x1!=x2) 25 { 26 cout << "0" << endl; 27 return 0; 28 } 29 // cout << x1 << " " << L << endl;//**** 30 while(x1<=L && x2<=L && x1!=x2) 31 { 32 if(x1<x2) 33 x1+=max(1ll, (x2-x1)/a1)*a1; 34 else if(x1>x2) 35 x2+=max(1ll, (x1-x2)/a2)*a2; 36 } 37 // cout << x1 << "**" << x2 << endl;//**** 38 if(x1==x2&&x1>=R&&x1<=L) 39 cout << (L-x1)/(a1/get_gcd(a1, a2)*a2)+1 << endl; 40 else cout << "0" << endl; 41 return 0; 42 }
据说这题应该用拓展欧里几德解。。可惜我看了好久也没弄懂。。诶。。继续看吧。。。。