Problem G. Generators

Input file: generators.in

Output file: generators.out
Little Roman is studying linear congruential generators — one of the oldest and best known pseudorandom number generator algorithms. Linear congruential generator (LCG) starts with a non-negative integer number x0 also known as seed and produces an infinite sequence of non-negative integer numbers xi (0 ≤ xi < c) which are given by the following recurrence relation:
xi+1 = (axi + b) mod c
here a, b, and c are non-negative integer numbers and 0 ≤ x0 < c. Roman is curious about relations between sequences generated by different LCGs. In particular, he has n different LCGs with parameters a(j), b(j), and c(j) for 1 ≤ j ≤ n, where the j-th LCG is generating a sequence x(j) i . He wants to pick one number from each of the sequences generated by each LCG so that the sum of the numbers is the maximum one, but is not divisible by the given integer number k. Formally, Roman wants to find integer numbers tj ≥ 0 for 1 ≤ j ≤ n to maximize s =Pn j=1 x(j) tj subject to constraint that s mod k 6= 0. Input The first line of the input file contains two integer numbers n and k (1 ≤ n ≤ 10000, 1 ≤ k ≤ 109). The following n lines describe LCGs. Each line contains four integer numbers x(j) 0 , a(j), b(j), and c(j) (0 ≤ a(j),b(j) ≤ 1000, 0 ≤ x(j) 0 < c(j) ≤ 1000). Output If Roman’s problem has a solution, then write on the first line of the output file a single integer s — the maximum sum not divisible by k, followed on the next line by n integer numbers tj (0 ≤ tj ≤ 109) specifying some solution with this sum. Otherwise, write to the output file a single line with the number −1.
Sample input and output
2 3

1 1 1 6

2 4 0 5

8

4

1

2 2

0 7 2 8

2 5 0 6

-1

In the first example, one LCG is generating a sequence 1, 2, 3, 4, 5, 0, 1, 2, ..., while the other LCG a sequence 2, 3, 2, 3, 2, ....

In the second example, one LCG is generating a sequence 0, 2, 0, 2, 0, ..., while the other LCG a sequence 2, 4, 2, 4, 2, ....

题目中的xi+1是指xi的下一项。

第四场训练赛的题,是欧洲赛的题,比赛地址:传送门

题意:n行,每行x,a,b,c,推公式,每一行在这一行当中取一个数使得他们的和最大且不能被k整除,如果不存在输出-1。

题解:记录第一大和第二大的数,并且第二大的数不能被k整除,求出所有行中与第一个数相差最小的这个第二个数。如果最大的数的所有总和被k整除,就用这个数去换,记得用数组记录选取的数的下标。鸽笼原理最多c个数,如果算过就可以跳出,用动态数组记录。

#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <vector>
using namespace std;
const int maxn=1e6+;
const int mm=;
bool vis[maxn];
int f1[maxn],f2[maxn];
int main()
{
freopen("generators.in","r",stdin);
freopen("generators.out","w",stdout);
int n,k,ans=,tmp=mm,flag=-;
scanf("%d%d",&n,&k);
for(int i=; i<n; i++)
{ int x,a,b,c;
scanf("%d%d%d%d",&x,&a,&b,&c);
for(int j=; j<c; j++)
vis[j]=;
vector <int> data;
for(int j=; j<c; j++)
{
if(vis[x]) break;
vis[x]=;
data.push_back(x);
x=(a*x+b)%c;
}
int max1=-,max1i=-;
for (int i = ; i < data.size(); i++)
{
if (data[i] > max1)
{
max1 = data[i];
max1i = i;
}
}
ans+=max1;
int max2=-,max2i=-;
int tmp2=max1%k;
for (int i = ; i < data.size(); i++)
{
if (data[i] > max2&&(data[i]%k)!=tmp2)
{
max2 = data[i];
max2i = i;
}
}
f1[i]=max1i;
f2[i]=max2i;
int minn=max1-max2;
if(max2i!=-&&minn<tmp)
{
tmp=minn;
flag=i;
}
//cout<<max1<<" "<<max2<<endl;
}
if(ans%k==&&flag==-)
{
cout<<-<<endl;
}
else if(ans%k==)
{
f1[flag]=f2[flag];
ans-=tmp;
cout<<ans<<endl;
for(int i=;i<n-;i++)
cout<<f1[i]<<" ";
cout<<f1[n-]<<endl;
}
else
{
cout<<ans<<endl;
for(int i=;i<n-;i++)
cout<<f1[i]<<" ";
cout<<f1[n-]<<endl;
}
return ;
}

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