【数论】【扩展欧几里得】Codeforces 710D Two Arithmetic Progressions
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题目链接:
http://codeforces.com/problemset/problem/710/D
题目大意:
两个等差数列a1x+b1和a2x+b2,求L到R区间内重叠的点有几个。
0 < a1, a2 ≤ 2·109, - 2·109 ≤ b1, b2, L, R ≤ 2·109, L ≤ R).
题目思路:
【数论】【扩展欧几里得】
据题意可得同余方程组 x=b1(mod a1) 即 x=k1*a1+b1
x=b2(mod a2) x=k2*a2+b2
化简,k1*a1=k2*a2+(b2-b1) 即 a1= (b2-b1)(mod a2)
于是只要求一个同余方程即可。令a=a1,b=a2,c=b2-b1。
扩展欧几里得求解x,再把x改为在L~R区间内的第一个通解,计算数量即可(每次增加lcm(a1,a2)答案+1)。
////by coolxxx//#include<bits/stdc++.h>#include<iostream>#include<algorithm>#include<string>#include<iomanip>#include<map>#include<memory.h>#include<time.h>#include<stdio.h>#include<stdlib.h>#include<string.h>//#include<stdbool.h>#include<math.h>#define min(a,b) ((a)<(b)?(a):(b))#define max(a,b) ((a)>(b)?(a):(b))#define abs(a) ((a)>0?(a):(-(a)))#define lowbit(a) (a&(-a))#define sqr(a) ((a)*(a))#define swap(a,b) ((a)^=(b),(b)^=(a),(a)^=(b))#define mem(a,b) memset(a,b,sizeof(a))#define eps (1e-8)#define J 10#define mod 1000000007#define MAX 0x7f7f7f7f#define PI 3.14159265358979323#define N 20000004using namespace std;typedef long long LL;int cas,cass;int n,m,lll,ans;LL a1,a2,b1,b2,l,r;LL exgcd(LL a,LL b,LL &x,LL &y){if(!b){x=1,y=0;return a;}LL d=exgcd(b,a%b,y,x);y-=a/b*x;return d;}int main(){#ifndef ONLINE_JUDGE//freopen("1.txt","r",stdin);//freopen("2.txt","w",stdout);#endifint i,j,k;LL a,b,c,d,x,y,ny,lcm;//for(scanf("%d",&cas);cas;cas--)//for(scanf("%d",&cas),cass=1;cass<=cas;cass++)//while(~scanf("%s",s+1))while(~scanf("%I64d",&a1)){cin>>b1>>a2>>b2>>l>>r;l=max(l,b1);l=max(l,b2);d=exgcd(a1,a2,x,y);lcm=a1/d*a2;if((b2-b1)%d!=0){puts("0");continue;}a=a1/d;b=a2/d;c=(b2-b1)/d;d=exgcd(a,b,x,y);x=a1*(x*c)+b1;if(x>l)x=x-(x-l)/lcm*lcm;else x=x+(l-1-x+lcm)/lcm*lcm;if(x>r)puts("0");else printf("%I64d\n",(r-x+lcm)/lcm-(l-1-x+lcm)/lcm);}return 0;}/*////*/
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