ACPC 2013 J.Modified LCS-优快云博客

本文介绍了一个经典的计算机科学问题——最长公共子序列(LCS)问题的一个变种，并提供了一种解决该问题的方法。该问题涉及寻找两个特定整数序列的最长共同子序列的长度，其中序列由初始值、长度及增量定义。文章还提供了实现这一算法的C++代码示例。

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Modified LCS

Time Limit: 10000ms, Special Time Limit:25000ms, Memory Limit:65536KB

Total submit users: 29, Accepted users: 17

Problem 12831 : No special judgement

Problem description

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 find 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 first element in the sequence. Each element except the first 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 first line of the input will be a single integer T, the number of test cases T ranges in(1,100). 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 (N1,N2 ranges in(1,10^18) and F1,D1,F2,D2 ranges in(1,10^9)) representing the length of the first sequence, the first element in the first sequence, the incremental value of the first sequence, the length of the second sequence, the first 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
50

Problem Source

ACPC 2013

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hnu挂的比赛，比赛时候纠结于扩展欧几里得没做出来简直是罪恶（hnu支持%I64d注意一下）.......

代码如下：

#include<stdio.h>
#include<iostream>
using namespace std;
typedef long long LL;
LL gcd(LL a,LL b)
{
    return b == 0?a:gcd(b,a%b);
}
LL exgcd(LL a,LL b,LL &x,LL &y)
{
    if(b == 0)
    {
        x = 1;
        y = 0;
        return a;
    }
    LL g = exgcd(b,a%b,x,y);
    LL temp;
    temp = x;
    x = y;
    y = temp - a/b*y;
    return g;
}
int main()
{
    LL a,b,c,n1,f1,d1,n2,f2,d2;
    LL xx,yy;
    int T;
    scanf("%d",&T);
    while(T--)
    {
        scanf("%I64d%I64d%I64d%I64d%I64d%I64d",&n1,&f1,&d1,&n2,&f2,&d2);

        a = d1;
        b = d2;
        c = (f2-f1)-(d2-d1);
        LL g = gcd(a,b);

        exgcd(a,b,xx,yy);
        xx = (((xx%(b/g)*(c/g)%(b/g)))%(b/g)+(b/g))%(b/g);
        if(xx == 0) xx+=b/g;
        yy = (c - a*xx)/b;
        yy = -yy;

        LL sum1 = f1+(n1-1)*d1;
        LL sum2 = f2+(n2-1)*d2;
        LL sum = 0,sum_first = 0,dd = 0;
        LL ans = 0;

        dd = d1/gcd(d1,d2)*d2;
        sum = sum2<sum1?sum2:sum1;
        sum_first = d1*(xx-1)+f1;
        ans = ((sum-sum_first)/dd)+1;

        printf("%I64d\n",ans);
    }
    return 0;
}