NOIP201410 solution equation (c + +)

Equation of NOIP201410 Solution

Difficulty level: A; run time limit: 1000ms; operating space limit: 51200KB; code length limit: 2000000B

Question Description

Known polynomial equation: a0+a1*x+a2*x^2+a3*x^3+...+an*x^n=0 The integer solution of this equation within [1, M] (both N and m are positive integers).

Input

Enter a total n+2 line. The first line contains 2 integers n, m, separated by a space between each of the two integers. The next n+1 line contains an integer for each row, followed by a0,a1,a2...an.

Output

The number of integer solutions in the first row of the output equation in [1, M]. Next, each line of an integer, according to the order from small to large output equation in [1, M] an integer solution.

Input example

2 10 2 -3 1

Output example

2 1 2

Other Notes

For 100% of data, 0 < n≤100, ai≤10^10000, an≠0,m≤1000000.

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define M 110
using namespace Std;
typedef long Long LL;
const int prime[]={30011,11261,14843,19997,10007,21893};
int n,m,stack[1001001],top;
int Ans[110],tot;
ll A[m][6],f[30011][6];
inline ll F (int x,int j)
{
int i;
ll Re=0;
for (i=n;~i;i--)
Re= (Re*x+a[i][j])%prime[j];
return re;
}
inline void Input (int x)
{
static Char s[10100];
int i,j;
BOOL Flag=false;
scanf ("%s", s+1);
for (i=1;s[i];i++)
{
if (s[i]== '-')
Flag=true;
Else
for (j=0;j<6;j++)
A[x][j]= ((a[x][j]<<1) + (a[x][j]<<3) + s[i]-' 0 ')%prime[j];
}
if (flag)
for (j=0;j<6;j++)
A[X][J]=PRIME[J]-A[X][J];
}
int main ()
{

Freopen ("Equation.in", "R", stdin);

int i,j;
cin>>n>>m;
for (i=0;i<=n;i++)
Input (i);

for (j=0;j<6;j++)
for (i=0;i<prime[j];i++)
F[i][j]=f (I,J);

for (i=0;i<prime[0];i++)
{
if (f[i%prime[0]][0]==0)
Stack[++top]=i;
}

for (i=1;i<=top;i++)
{
if (stack[i]+prime[0]>m)
Break
STACK[++TOP]=STACK[I]+PRIME[0];
}

for (i=1;i<=top;i++)
{
if (stack[i]>m)
Break
for (j=1;j<6;j++)
if (F[stack[i]%prime[j]][j])
Break
if (j==6)
Ans[++tot]=stack[i];
}

cout<<tot<<endl;
for (i=1;i<=tot;i++)
printf ("%d\n", Ans[i]);
}

NOIP201410 solution equation (c + +)