LA 4119 - Always an integer

https://icpcarchive.ecs.baylor.edu/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&problem=2120

假设式子的最高幂次为k 则只需要测试1到k+1就可以了

原理见 刘汝佳的《算法竞赛入门经典入门指南》 P123

字符串需要自己处理，比较繁琐 一定要细心

代码：

#include <iostream>

#include <cstdio>

#include <string>

#include <cstring>

#include <cmath>

#include <algorithm>

#include <queue>



#define ll long long

#define lint long long

using namespace std;



const int N=100005;

ll a[N];

ll powerMod(ll x,ll y,ll M)

{//cout<<x<<" "<<y<<" "<<M<<endl;

    ll tmp=1;

    while(y)

    {

        if(y&1)

        tmp=tmp*x%M;

        x=x*x%M;

        y=y>>1;

    }

    //cout<<(tmp%M)<<endl;

    return tmp%M;

}

void func(string s)

{//cout<<s<<endl;

    if(s.size()==0) return;



    int m=s.find('n');

    if(m==-1)

    m=s.size();

    int type=(s[0]=='-')?-1:1;

    ll tmp=0;

    //int l=;cout<<l<<" "<<m<<endl;

    for(int i=(s[0]=='+'||s[0]=='-')?1:0;i<m;++i)

    {//cout<<i<<" "<<tmp<<endl;

        tmp=tmp*10+s[i]-'0';

    }

    if(tmp==0) tmp=1;

    tmp=tmp*type;

    m=s.find('n');//cout<<m<<endl;

    if(m==-1) a[0]=tmp;

    else

    {//cout<<"in1"<<endl;return ;

        int k=s.find('^');

        if(k==-1)

        a[1]=tmp;

        else

        {//cout<<"in2"<<endl;

            int x=0;

            for(int i=k+1;i<s.size();++i)

            {

                x=x*10+s[i]-'0';

            }

            //cout<<x<<" "<<tmp<<endl;

            a[x]=tmp;

        }

    }

    //cout<<"return"<<endl;

}

bool solve(string s)

{

    memset(a,0,sizeof(a));

    int k=s.find('/');

    ll M=0;

    for(int i=k+1;i<s.length();++i)

    M=M*10+s[i]-'0';

    string stmp;

    for(int i=(s[0]=='(')?1:0;i<k&&s[i]!=')';++i)

    {

        if(s[i]=='+'||s[i]=='-')

        {

            func(stmp);

            stmp.clear();

        }

        stmp.push_back(s[i]);

    }

    func(stmp);

    for(ll i=1;i<=101;++i)

    {//cout<<i<<endl;

        ll tmp=0;

        for(ll j=0;j<=100;++j)

        {

            //cout<<j<<" "<<a[j]<<endl;

            tmp=(tmp+a[j]*powerMod(i,j,M))%M;

        }

        if(tmp!=0)

        return false;

    }

    return true;

}

int main()

{

    //freopen("data.in","r",stdin);

    int ca=1;

    string s;

    while(cin>>s)

    {

        if(s==".")break;

        if(solve(s))

        printf("Case %d: Always an integer\n",ca++);

        else

        printf("Case %d: Not always an integer\n",ca++);

    }

    return 0;

}