worldfinals 2008 Always an integer

  利用差分序列  p(n) 为K 次多项式       p(0)p(1) p(2) .........p(k) 均整除D  其一阶 差分序列  dp(0),dp(1)....dp(n) 均能整除D，类推可知  d^kp(0),d^kp(1).....可整除D，且差分序列最后  变为常数

  

#include<cstdio>#include<cstring>#include<queue>#include<vector>#include<iostream>#include<string>#include<sstream>#include<cctype>#include<set>#include<stack>#include<functional>#include<memory>#include<deque>#include<list>#include<cmath>#include<fstream>#include<cstdlib>#include<climits>#include<iomanip>#include<cstring>#include<memory>#include<bitset>#include<algorithm>using namespace std;typedef long long ll;typedef vector<int>VI;typedef vector<VI>VVI;typedef vector<VVI>VVVI;const int maxn = 711;const int mod = 1000000007;template<class T> T gcd(T a,T b){ return b?gcd(b,a%b):a;}template<class T> T lcm(T a,T b){ return a/gcd(a,b)*b;}char buf[maxn<<2];ll POW(ll a,ll b,ll c){ll res=1;while(b){if(b&1) res=res*a%c;a=a*a%c;b>>=1;}return res;}ll E[maxn],p[maxn],D;int cnt=0;bool cal(int k,int n){ll res=0,t;int i;for(i=0;i<cnt;++i){t=E[i]%D;t=t*POW(k,p[i],D)%D;res=(res+t)%D;}return res==0;}bool ok(char ch){return ch=='+'||ch=='-';}int main(){ios::sync_with_stdio(false);int cas=0,i,j,beg;while(true){cas++;cin>>buf;if(buf[0]=='.') break;char ch;cnt=0;memset(p,0,sizeof(p));int sign,n,len=strlen(buf);if(buf[1]!='-') beg=0,buf[0]='+';else beg=1;for(i=beg;i<len;){if(buf[i]=='/'){D=0;for(j=i+1;j<len;++j){D=D*10+(buf[j]-'0');}break;}if(ok(buf[i])){sign=1;if(buf[i]=='-') sign=-1;E[cnt]=0;for(j=i+1;buf[j]!='n'&&!ok(buf[j])&&buf[j]!=')';++j){E[cnt]=E[cnt]*10+(buf[j]-'0');}E[cnt]*=sign;if(!E[cnt])E[cnt]=sign;i=j;p[cnt]=0;if(buf[i]=='n'){if(buf[i+1]=='^'){p[cnt]=0;for(j=i+2;!ok(buf[j])&&buf[j]!=')';++j){p[cnt]=p[cnt]*10+(buf[j]-'0');}}elsej=i+1,p[cnt]=1;}i=j;if(buf[i]==')')++i;cnt++;}}bool flag=true;for(n=0;n<=p[0];++n){flag&=cal(n,cnt);if(!flag)break;}cout<<"Case "<<cas<<": ";if(flag)cout<<"Always an integer"<<endl;elsecout<<"Not always an integer"<<endl;}    return 0;}