luogu P2312 解方程

二次联通门 : luogu P2312 解方程

 

 

 

 

/*
    luogu P2312 解方程

    知道正解的我眼泪掉下来
    原来模一模就能过。。
    当时咋就不知道写呢。。。


 */
#include <cstdio>
#include <iostream>
#include <cstring>

#define Max 105
#define L 5
int Mod[L] = {9973,9931,9941,9949,9967};
int a[L][Max], cal[L][Max * 100], pre[L][Max];
int Answer[Max * 10000], C;

void read (int &now)
{
    register char c = getchar ();
    for (now = 0; !isdigit (c); c = getchar ());
    for (; isdigit (c); now = now * 10 + c - '0', c = getchar ());
}

char line[Max * 100];

inline bool Judge (int now)
{
    for (int i = 0; i < L; ++ i)
        if (cal[i][now % Mod[i]]) return false;
    return true;
}

int Main ()
{
    freopen ("equation.in", "r", stdin);
    freopen ("equation.ans", "w", stdout);

    int N, M, Len; register int i, j, k; bool temp = false;
    read (N), read (M);
    for (i = 0; i <= N; ++ i)
    {
        temp = false; scanf ("%s", line); Len = strlen (line);
        if (line[0] == '-')
            temp = true;
        else if (isdigit (line[0]))
            for (j = 0; j < L; ++ j)
                a[j][i] = line[0] - '0';
        for (j = 0; j <    L; ++ j)
            for (k = 1; k < Len; ++ k)
                a[j][i] = (a[j][i] * 10 + line[k] - '0') % Mod[j];
        if (temp) for (j = 0; j < L; ++ j) a[j][i] = -a[j][i];
    }
    int res = 0;
    for (i = 0; i < L; ++ i)
        for (j = 1; j < Mod[i]; ++ j)
        {
            pre[i][0] = 1;
            for (k = 1; k <= N; ++ k) 
                pre[i][k] = (pre[i][k - 1] * j) % Mod[i];
            res = 0;
            for (k = 0; k <= N; ++ k)
                res = (res + a[i][k] * pre[i][k]) % Mod[i];
            cal[i][j] = res < 0 ? res + Mod[i] : res;
        }
        
    for (i = 1; i <= M; ++ i)
        if (Judge (i)) Answer[++ C] = i;
    printf ("%d\n", C);
    for (i = 1; i <= C; ++ i)
        printf ("%d\n", Answer[i]);

    return 0;
}

int ZlycerQan = Main ();
int main (int argc, char *argv[]) {;}