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文章目录
- F.Longest Common Subsequence
F.Longest Common Subsequence
链接：https://ac.nowcoder.com/acm/contest/33193/F
来源：牛客网
题目描述
Given sequence ss of length nn and sequence tt of length mm, find the length of the longest common subsequence of ss and tt.
输入描述:
There are multiple test cases. The first line of input contains an integer TT(1\le T\le 10^31≤T≤10
3
), the number of test cases.For each test case:
The only line contains 77 integers, nn, mm, pp, xx, aa, bb, cc (1\le n1≤n, m\le 10^6m≤10
6
, 0\le x0≤x, aa, bb, c<p\le 10^9c<p≤10
9
). nn is the length of ss, mm is the length of tt.To avoid large input, you should generate the sequences as follows:
For each i=1i=1, 22, \cdots⋯, nn in order, update xx to (ax^2+bx+c)\bmod p(ax
2
+bx+c)modp, and then set s_is
i
to xx. And then, for each i=1i=1, 22, \cdots⋯, mm in order, update xx to (ax^2+bx+c)\bmod p(ax
2
+bx+c)modp, and then set t_it
i
to xx.
It is guaranteed that the sum of nn and the sum of mm over all test cases does not exceed 10^610
6
.
输出描述:
For each test case:
Output an integer – the length of the longest common subsequence of ss and tt, in one line.
示例1
输入
复制
2
4 3 1024 1 1 1 1
3 4 1024 0 0 0 0
输出
复制
0
3
说明
In the first sample, s=[3,13,183,905]s=[3,13,183,905] and t=[731,565,303]t=[731,565,303].
In the second sample, s=[0,0,0]s=[0,0,0] and t=[0,0,0,0]t=[0,0,0,0].
题意：
- 给出n, m, p, x, a, b, c。 令
。
让长为n的序列s，s[1]=f(x), s[2]=f(s[1]), s[3]=f(s[2])，依次递推。
让长为m的序列t，t[1]=f(s[n]), t[2] = f(t[1]), t[3] = f(t[2])，依次类推。 - 最后求序列s和序列t的LCS（最长公共子序列）。
思路：
- 因为要%p，所以可以发现，到后面这两个序列都是循环的。换句话说，如果两个序列中有相同数字，那么后面的数字一定全一样。
- 开个map记录下序列s中每个数第一次出现的位置。对于t进行遍历的时候，找到这个数的位置，将序列t[j]与s[i]对上后，不难发现 [ t[j], m ] 与 [ s[i], n ] 这两个区间一定是一样的，因为都是往后无限循环，取个min就是LCS的大小。遍历一遍维护最大值即可。
#include<bits/stdc++.h>
using namespace std;
#define IOS ios::sync_with_stdio(0), cin.tie(0),cout.tie(0)
typedef long long LL;
int main(){
LL T; cin>>T;
while(T--){
LL n, m, p, x, a, b, c; cin>>n>>m>>p>>x>>a>>b>>c;
unordered_map<LL,LL>mp; //每个数最早出现的位置
for(LL i = 1; i <= n; i++){
x = (a*x%p*x+b*x+c)%p;
if(!mp.count(x))mp[x] = i;
}
LL ans = 0;
for(LL i = 1; i <= m; i++){
x = (a*x%p*x+b*x+c)%p;
if(mp.count(x)){
//LCS可选 [mp[x],n] 或 [i,m]
ans = max(ans, min(n-mp[x], m-i)+1);
}
}
cout<<ans<<"\n";
}
return 0;
}
