uva1069 Always an integer【解法二】

Combinatorics is a branch of mathematics chie y concerned with
counting discrete objects. For instance, how many ways can you pick
two people out of a crowd of n people? Into how many regions can you
divide a circular disk by connecting n points on its boundary with one
an- other? How many cubes are in a pyramid with square layers ranging
from 1  1 to n  n cubes? Many questions like these have answers that
can be reduced to simple polynomials in n . The answer to the rst
question above is n ( n

解法一【随机化】见【这里】。
设最高次数为k，只需要代入n=1,2,…,k+1验证。
证明如下：
k=0，显然只需要带入p(1)。
k>=1，可以发现首项和相邻两项的差都能整除d就是整个数列能整除d的充要条件。差分数列f(n)=p(n+1)-p(n)是一个k-1次项式，根据归纳法可以知道需要需要保证f(1),f(2),..,f(k)全能整除d，再加上p(1)能整除d，也就是p(1),p(2),…,p(k+1)能整除d。

#include
#include
#include
#define LL long long
const int T=10000;
LL a[110],t[110],d,r[10010];
char s[100010];
int n;
bool init()
{
    int x,f,p=0;
    char c;
    scanf("%s",s);
    c=s[p++];
    if (c=='.') return 0;
    n=0;
    c=s[p++];
    while (c!=')')
    {
        n++;
        if (c=='-')
        {
            f=-1;
            c=s[p++];
        }
        else f=1;
        if (c=='+') c=s[p++];
        x=0;
        while (c>='0'&&c<='9')
        {
            x=x*10+c-'0';
            c=s[p++];
        }
        if (!x) x=1;
        x*=f;
        a[n]=x;
        if (c!='n')
        {
            t[n]=0;
            continue;
        }
        c=s[p++];
        x=0;
        if (c=='^')
        {
            c=s[p++];
            while (c>='0'&&c<='9')
            {
                x=x*10+c-'0';
                c=s[p++];
            }
        }
        else x=1;
        t[n]=x;
    }
    c=s[p++];
    c=s[p++];
    x=0;
    while (c>='0'&&c<='9')
    {
        x=x*10+c-'0';
        c=s[p++];
    }
    d=x;
    return 1;
}
LL pow(LL base,LL p)
{
    LL ret=1;
    while (p)
    {
        if (p&1) ret=ret*base%d;
        base=base*base%d;
        p>>=1;
    }
    return ret;
}
bool solve()
{
    int i,j;
    LL x;
    for (i=1;i<=T;i++)
    {
        x=0;
        for (j=1;j<=n;j++)
          x=((x+pow(r[i],t[j])*a[j]%d)%d+d)%d;
        if (x) return 0;
    }
    return 1;
}
void make()
{
    int i;
    srand(123);
    for (i=1;i<=T;i++)
      r[i]=rand();
}
int main()
{
    int K=0;
    make();
    while (init())
    {
        printf("Case %d: ",++K);
        if (solve()) printf("Always an integer\n");
        else printf("Not always an integer\n");
    }
}