poj3243

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解高次同余方程的应用题
#include<cstdio>#include<cstring>#include<string>#include<algorithm>#include<cmath>#define LL long longusing namespace std;const int maxn = 65535;struct hash{    int a,b,next;} Hash[maxn << 1];int flg[maxn + 66];int top,idx;void ins(int a,int b){    int k = b & maxn;    if(flg[k] != idx)    {        flg[k] = idx;        Hash[k].next = -1;        Hash[k].a = a;        Hash[k].b = b;        return ;    }    while(Hash[k].next != -1)    {        if(Hash[k].b == b) return ;        k = Hash[k].next;    }    Hash[k].next = ++ top;    Hash[top].next = -1;    Hash[top].a = a;    Hash[top].b = b;}int find(int b){    int k = b & maxn;    if(flg[k] != idx) return -1;    while(k != -1)    {        if(Hash[k].b == b) return Hash[k].a;        k = Hash[k].next;    }    return -1;}int gcd(int a,int b){    return b?gcd(b,a%b):a;}int ext_gcd(int a,int b,int& x,int& y){    int t,ret;    if (!b)    {        x=1,y=0;        return a;    }    ret=ext_gcd(b,a%b,x,y);    t=x,x=y,y=t-a/b*y;    return ret;}int Inval(int a,int b,int n){    int x,y,e;    ext_gcd(a,n,x,y);    e=(LL)x*b%n;    return e<0?e+n:e;}int pow_mod(LL a,int b,int c){    LL ret=1%c;    a%=c;    while(b)    {        if(b&1)ret=ret*a%c;        a=a*a%c;        b>>=1;    }    return ret;}int BabyStep(int A,int B,int C){    top = maxn;    ++ idx;    LL buf=1%C,D=buf,K;    int i,d=0,tmp;    for(i=0; i<=100; buf=buf*A%C,++i)if(buf==B)return i;    while((tmp=gcd(A,C))!=1)    {        if(B%tmp)return -1;        ++d;        C/=tmp;        B/=tmp;        D=D*A/tmp%C;    }    int M=(int)ceil(sqrt((double)C));    for(buf=1%C,i=0; i<=M; buf=buf*A%C,++i)ins(i,buf);    for(i=0,K=pow_mod((LL)A,M,C); i<=M; D=D*K%C,++i)    {        tmp=Inval((int)D,B,C);        int w ;        if(tmp>=0&&(w = find(tmp)) != -1)return i*M+w+d;    }    return -1;}int main(){    int a,b,c;    while(~scanf("%d%d%d",&a,&c,&b)&&a&&b&&c)    {        int temp = BabyStep(a,b,c);        if(temp<0)            printf("No Solution\n");        else            printf("%d\n",temp);    }    return 0;}
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