NOIP2014D2T3 is good at solving the equation Hash big method.

Given the high-order equation an * x ^ n... a1 * x ^ 1a0 * x ^ 00 calculate the number of integers in the range [1, m] ai10 ^ 10000, And the m100W is too lazy to be precise. The longdouble written in the test room is messy ...... 30 minutes, 50 minutes, and a day, QAQ. After I finally got all sorts of unofficial data, I had to kneel down on the earth, hold flowers, and look up at the sky and shout: hash!

Given the high-order equation an * x ^ n... a1 * x ^ 1a0 * x ^ 0 = 0 how many integers are there in the range [1, m]? ai = 10 ^ 10000, m = 100 W too lazy and high-precision, long double in the test room ...... 30 minutes, 50 minutes, and a day, QAQ. After I finally got all sorts of unofficial data, I had to kneel down on the earth, hold flowers, and look up at the sky and shout: hash!

Given the high-order equation an * x ^ n +... + a1 * x ^ 1 + a0 * x ^ 0 = 0 evaluate the number of integer roots in the range [1, m]

Ai <= 10 ^ 10000, m <= 100 W

Too lazy and high-precision, long double in the test room ...... 30 minutes base, 50 minutes top day, QAQ

After I finally got all sorts of unofficial data, I had to kneel down on the earth, hold flowers in my hand, and look up at the sky and shout: Hash big way!

First, Abel told us five or more times before June 200 that there was no root equation, so we could only enumerate 1 ~ M: 100 W.

Then the accuracy of plus bits does not need to be calculated ...... What should we do? QAQ

Hash is good!

Make f (x) = an * x ^ n +... + a1 * x ^ 1 + a0 * x ^ 0 it is easy to know if f (x) = 0 then f (x) mod p = 0

If f (x) mod p = 0, then we can basically conclude that f (x) = 0 p is very reliable, and the collision rate is very low.

So we can modularize all the ai and then verify each solution O (n ).

In this case, if the O (m * n) can get 70 points p, it will not fail.

So what about 100 points?

Hash is good!

We can easily find f (x + p) mod p = f (x) mod p

Therefore, we select a smaller p and pre-process 0 ~ All the f (x) values of the p-1st, And the modulo exceeding the p value can be obtained.

But p is not enough for the conference!

So we select a few more p to get them one time. mod. Only one value is counted as the solution after all p modulo values are 0.

Hash is good! Hash Killer III is proving the correctness of this algorithm! Long live hash! Hash competition height! Hash!

Time complexity O (n Σ p + m) Where Σ p is the sum of all the selected prime numbers. Generally, we can choose a prime number of about 1 W.

I don't know why, no matter how much points I got in the test room, I just need to come back and cut the question. Even if I am mentally AC, I am 0.0 ......

#include
 
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    #include#define M 110using namespace std;typedef long long ll;const int prime[]={10007,11261,14843,19997,21893};int n,m,stack[1001001],top;ll a[M][5],f[21893][5];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<5;j++)                a[x][j]=( (a[x][j]<<1) + (a[x][j]<<3) + s[i]-'0' )%prime[j];    }    if(flag)         for(j=0;j<5;j++)            a[x][j]=prime[j]-a[x][j];}int main(){    int i,j;    cin>>n>>m;    for(i=0;i<=n;i++)        Input(i);         for(j=0;j<5;j++)        for(i=0;i