[NOIP 2014] 解方程

[题目链接]

         https://www.lydsy.com/JudgeOnline/problem.php?id=3751

[算法]

        对于每个x ， 将方程左边对一个质数取模

        时间复杂度 ： O(NM)

[代码]

         

#include<bits/stdc++.h>
using namespace std;
#define MAXN 300010
#define RG register
const int P = 998244353;
typedef long long LL;

int n , m , ans , cnt;
int a[MAXN] , res[MAXN];

template <typename T> inline void chkmax(T &x , T y) { x = max(x , y); }
template <typename T> inline void chkmin(T &x , T y) { x = min(x , y); }
template <typename T> inline void read(T &x)
{
    T f = 1; x = 0;
    char c = getchar();
    for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
    for (; isdigit(c); c = getchar()) x = (1LL * x * 10 + c - '0') % P;
    x *= f;
    x = (x + P) % P;
}
inline bool ok(int x)
{
    int ret = 0 , tmp = 1;
    for (RG int i = 0; i <= n; i++)
    {
        ret = (ret + 1LL * a[i] * tmp) % P;
        tmp = 1LL * tmp * x % P;    
    }    
    if (!ret) return true;
    else return false;
}

int main()
{

    read(n); read(m);
    for (RG int i = 0; i <= n; i++) read(a[i]);
    for (RG int i = 1; i <= m; i++)
    {
        if (ok(i))
        {
            ++ans;
            res[++cnt] = i;
        }        
    }
    printf("%d\n" , ans);
    for (RG int i = 1; i <= cnt; i++) printf("%d\n" , res[i]);
    
    return 0;
    
}