【hdu2815-Mod Tree】高次同余方程-拓展BadyStepGaintStep
http://acm.hdu.edu.cn/showproblem.php?pid=2815
题意：裸题。。。
关于拓展BSGS的详细解释我写了一篇博文：http://www.cnblogs.com/KonjakJuruo/p/5178600.html
题解：又有一个坑，就是N>=P的时候输出无解。
1 #include<cstdio> 2 #include<cstdlib> 3 #include<cstring> 4 #include<iostream> 5 #include<cmath> 6 #include<algorithm> 7 using namespace std; 8 9 typedef long long LL; 10 const int SIZE=40000; 11 LL K,P,N; 12 int bl; 13 struct node{ 14 LL d,id; 15 }bit[SIZE]; 16 17 bool cmp(node x,node y){ 18 if(x.d==y.d) return x.id<y.id; 19 return x.d<y.d; 20 } 21 22 LL exgcd(LL a,LL b,LL &x,LL &y) 23 { 24 if(b==0) {x=1,y=0;return a;} 25 LL tx,ty; 26 LL d=exgcd(b,a%b,tx,ty); 27 x=ty;y=tx-(a/b)*ty; 28 return d; 29 } 30 31 LL find(LL x) 32 { 33 int l=0,r=bl; 34 while(l<=r) 35 { 36 int mid=(l+r)>>1; 37 if(bit[mid].d==x) return bit[mid].id; 38 if(bit[mid].d<x) l=mid+1; 39 if(bit[mid].d>x) r=mid-1; 40 } 41 return -1; 42 } 43 44 LL exBSGS() 45 { 46 if(N>=P) return -1; 47 LL t,m,a,b,c,g,k,x,y,add,pm,am; 48 k=1;a=K;b=N;c=P;add=0;t=1; 49 for(int i=0;i<=100;i++)//在约分前 50 { 51 if(t%c==b) return i; 52 t=t*a%c; 53 } 54 while((g=exgcd(K,c,x,y)) != 1) 55 { 56 k=(k*K/g)%c; 57 c/=g;//不是mod 58 if(b%g) return -1; 59 b/=g;add++; 60 } 61 m=(LL)(ceil((double)sqrt((double)c))); 62 pm=1%c;bit[0].d=k%c;bit[0].id=0; 63 for(int i=1;i<=m;i++) 64 { 65 bit[i].d=bit[i-1].d*a%c; 66 bit[i].id=i; 67 pm=pm*a%c; 68 } 69 sort(bit,bit+1+m,cmp); 70 bl=0; 71 for(int i=1;i<=m;i++) 72 { 73 if(bit[i].d!=bit[bl].d) bit[++bl]=bit[i]; 74 } 75 exgcd(pm,c,x,y); 76 am=x%c+c;//x 77 t=b%c; 78 for(int i=0;i<=m;i++) 79 { 80 x=find(t); 81 if(x!=-1) return i*m+x+add; 82 t=t*am%c; 83 } 84 return -1; 85 } 86 87 int main() 88 { 89 // freopen("a.in","r",stdin); 90 // freopen("a.out","w",stdout); 91 while(scanf("%I64d%I64d%I64d",&K,&P,&N)!=EOF) 92 { 93 LL ans=exBSGS(); 94 if(ans==-1) printf("Orz,I can’t find D!\n"); 95 else printf("%I64d\n",ans); 96 } 97 return 0; 98 }
