LCS stands for longest common subsequence, and it is a well known problem. A sequence in this

problem means a list of integers, and a sequence X is considered a subsequence of another sequence Y ,

when the sequence X can be obtained by deleting zero or more elements from the sequence Y without

changing the order of the remaining elements.

In this problem you are given two sequences and your task is to nd the length of the longest

sequence which is a subsequence of both the given sequences.

You are not given the sequences themselves. For each sequence you are given three integers N, F

and D, where N is the length of the sequence, F is the rst element in the sequence. Each element

except the rst element is greater than the element before it by D.

For example N = 5, F = 3 and D = 4 represents the following sequence: [3, 7, 11, 15, 19].

There will be at least one integer which belongs to both sequences and it is not greater than

1,000,000.

Input

Your program will be tested on one or more test cases. The rst line of the input will be a single integer

T, the number of test cases (1 T 100). Followed by the test cases, each test case is described in one

line which contains 6 integers separated by a single space N1 F1 D1 N2 F2 D2 (1 N1;N2 1018)

and (1 F1;D1; F2;D2 109) representing the length of the rst sequence, the rst element in the

rst sequence, the incremental value of the rst sequence, the length of the second sequence, the rst

element in the second sequence and the incremental value of the second sequence, respectively.

Output

For each test case, print a single line which contains a single integer representing the length of the

longest common subsequence between the given two sequences.

Sample Input

3

5 3 4 15 3 1

10 2 2 7 3 3

100 1 1 100 1 2

Sample Output

4

3

 

这条题是组队排位的中下题。小邪卡了很久。

解法其实就是用扩展欧几里得算法求出第一个出现的相等的位置。

f1  + (x-1)*d1 = f2 + (y1 -1 )*d2;

除去前面的位置，接下来就是算两个等差序列有多少段出现LCS，然后取较小那个就可以了

 

 

#include
     
      using namespace std;typedef long long LL;void e_gcd(LL a,LL b,LL &x,LL &y,LL &d){    if( !b ){ d = a,x = 1 ,y = 0; return ;}    e_gcd(b,a%b,y,x,d);    y -= x * (a / b);}int main(){    int _;    LL k1,k2,c;    LL x,y,L1,L2,R1,R2,ML,MR,d;    LL f1,f2,d1,d2,n1,n2;    scanf("%d",&_);    while(_--){        LL ans=0;        scanf("%lld%lld%lld%lld%lld%lld",&n1,&f1,&d1,&n2,&f2,&d2);        c = f1 -f2;        d = __gcd(d1,d2);        if( c % d ){puts("0");continue;}        e_gcd(d1,d2,x,y,d);        k1 = -x * (c/d);        k2 =  y * (c/d);        d1 /= d , d2 /= d;        L1 = ceil( (-k1*1.0)/d2 );        L2 = ceil( (-k2*1.0)/d1 );        R1 = floor( (n1-k1) * 1.0 / d2);        R2 = floor( (n2-k2) * 1.0 / d1);        if( (n1 - k1) %d2 ==0  )R1 --;        if( (n2 - k2) %d1 ==0  )R2 --;        ML = max(L1,L2);        MR = min(R1,R2);        ans = max(0LL,MR - ML +1 );        printf("%lld\n",ans);    }    return 0;}