noip2014解题报告

D2T3：

秦九韶算法

#include<cstdio>
#include<cstring>
#include<iostream>
#include<cstdlib>
#include<algorithm>
#define ll long long
using namespace std;
const int maxn=1e2+5;
const int maxm=1e7+5;
const ll P1=999653;
const ll P2=673313;
const ll P3=999983;
ll a[maxn][5],ans[maxm];
int n,m;
inline void read(int i){
    a[i][1]=a[i][2]=a[i][3]=0ll;
    ll b=1;
    char x=getchar();
    while(x<'0'||'9'<x){
        if(x=='-')b=-1;
        x=getchar();
    }
    while('0'<=x&&x<='9'){
    a[i][1]=( (a[i][1]<<1) + (a[i][1]<<3) + x-'0')%P1;    
    a[i][2]=( (a[i][2]<<1) + (a[i][2]<<3) + x-'0')%P2;    
    a[i][3]=( (a[i][3]<<1) + (a[i][3]<<3) + x-'0')%P3;    
    x=getchar();
    }
    a[i][1]*=b;
    a[i][2]*=b;
    a[i][3]*=b;
}
inline int read2(){
    int a=0;bool b=1;char x=getchar();
    while(x<'0'||'9'<x){
        if(x=='-')b=0;
        x=getchar();
    }
    while('0'<=x&&x<='9'){
        a=(a<<1)+(a<<3)+x-'0';
        x=getchar();
    }
    return b ? a : -a ;
}
int main()
{
    //freopen("equation.in","r",stdin);
    //freopen("equation.out","w",stdout);
    n=read2();m=read2();
    for(int i=0;i<=n;i++){
        read(i);
    } 
    for(int i=1;i<=m;i++){
        ll k1=0,k2=0,k3=0;
        for(int j=n;j>=0;j--){
            k1=(k1*i+a[j][1])%P1;
            k2=(k2*i+a[j][2])%P2;
            k3=(k3*i+a[j][3])%P3;
        }
        if( (!k1) && (!k2) && (!k3) )ans[ ++ans[0] ]=i;
    }
    for(int i=0;i<=ans[0];i++){
    printf("%lld\n",ans[i]);
    }
    return 0;
}
/*10457is prime
10459is prime
10463is prime*/