Description has only two Sentences(hdu3307)

Description has only two Sentences

Time Limit: 3000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 1108    Accepted Submission(s): 345


Problem Description
an = X*an-1 + Y and Y mod (X-1) = 0.
Your task is to calculate the smallest positive integer k that ak mod a0 = 0.
 

 

Input
Each line will contain only three integers X, Y, a0 ( 1 < X < 231, 0 <= Y < 263, 0 < a0 < 231).
 

 

Output
For each case, output the answer in one line, if there is no such k, output "Impossible!".
 

 

Sample Input
2 0 9
 

 

Sample Output
1
 

 

Author
WhereIsHeroFrom
 思路：baby_step；
先构造等比数列，a[n]=X*a[n-1]+Y;那么（a[n]+k）=X（a[n-1]+k）;那么展开求得k=（Y）/（X-1）；那么a[n]+k=（a[0]+k）*(x)^n;
然后a[n]=(a[0]+k)*(x)^n-k;那么要求最小的n满足a[n]%a[0]=0;那么就是求k*x^n%(a[0]) = k%a[0],那么这个用扩展baby_step；求下就可以了，复杂度（sqrt（a[0]）*log(a[0])）;
  1 #include<stdio.h>
  2 #include<algorithm>
  3 #include<iostream>
  4 #include<string.h>
  5 #include<queue>
  6 #include<set>
  7 #include<math.h>
  8 #include<map>
  9 using namespace std;
 10 typedef long long LL;
 11 typedef struct node
 12 {
 13         LL x;
 14         LL id;
 15 } ss;
 16 ss ans[600000];
 17 ss bns[600000];
 18 pair<LL,LL>exgcd(LL n,LL m);
 19 LL gcd(LL n,LL m);
 20 LL quick(LL n,LL m,LL mod);
 21 bool cmp(node p,node q)
 22 {
 23         if(p.x==q.x)
 24                 return p.id<q.id;
 25         else return p.x<q.x;
 26 }
 27 LL er(int n,int m,LL k);
 28 int main(void)
 29 {
 30         LL x,y,a;
 31         while(scanf("%lld %lld %lld",&x,&y,&a)!=EOF)
 32         {
 33                 int kp = 0;
 34                 LL t = y/(x-1);//printf("%lld\n",t);
 35                 LL tt  = t%a;
 36                 if(t%a==0)
 37                 {
 38                         printf("1\n");
 39                 }
 40                 else
 41                 {
 42                         t%=a;
 43                         int flag = 0;
 44                         LL cnt = 0;
 45                         LL as = 1;
 46                         while(true)
 47                         {
 48                                 LL gc = gcd(x,a);
 49                                 if(gc == 1)
 50                                         break;
 51                                 else if(t%gc==0)
 52                                 {
 53                                         cnt++;
 54                                         a/=gc;
 55                                         as*=x/gc;
 56                                         t/=gc;
 57                                         as%=a;
 58                                 }
 59                                 else if(t%gc)
 60                                 {
 61                                         kp = 1;
 62                                         break;
 63                                 }
 64                                 if(t*as%gc==t)
 65                                 {
 66                                         flag = 1;
 67                                         break;
 68                                 }
 69                         }
 70                         if(kp)printf("Impossible!\n");
 71                         else if(flag)
 72                                 printf("%lld\n",cnt+1);
 73                         else
 74                         {
 75                                 LL v = sqrt(1.0*a);
 76                                 pair<LL,LL>acc = exgcd(x,a);
 77                                 LL  xx = quick(x,v,a);
 78                                 int i;//printf("%lld\n",xx);
 79                                 acc.first = (acc.first%a+a)%a;
 80                                 LL sum = t*acc.first%a;
 81                                 int cn = 0;
 82                                 for(i = 1; i <= v; i++)
 83                                 {
 84                                         ans[cn].x= sum%a;
 85                                         ans[cn].id = i;
 86                                         cn++;
 87                                         sum = sum*acc.first%a;
 88                                 }
 89                                 sort(ans,ans+cn,cmp);
 90 
 91                                 bns[0]=ans[0];
 92                                 LL cc = ans[0].x;
 93                                 int ac = 1;
 94                                 for(i = 1; i < cn; i++)
 95                                 {
 96                                         if(ans[i].x!=cc)
 97                                         {
 98                                                 cc = ans[i].x;
 99                                                 bns[ac].x = ans[i].x;
100                                                 bns[ac].id = ans[i].id;
101                                                 ac++;
102                                         }
103                                 }
104                                 LL akk = as*tt%a;
105                                 LL idd;//printf("%lld\n",bns[2].x);
106                                 for(i = 0; i <= v; i++)
107                                 {
108                                         idd = er(0,ac-1,akk);
109                                         if(idd!=-1)
110                                                 break;
111                                         akk = akk*xx%a;
112                                 }
113                                 if(i==v+1)
114                                 {
115                                         printf("Impossible!\n");
116                                 }
117                                 else
118                                 {
119                                         printf("%lld\n",cnt+i*v+idd);
120                                 }
121                         }
122                 }
123         }
124         return 0;
125 }
126 pair<LL,LL>exgcd(LL n,LL m)
127 {
128         if(m==0)
129                 return make_pair(1,0);
130         else
131         {
132                 pair<LL,LL>ak = exgcd(m,n%m);
133                 return make_pair(ak.second,ak.first-(n/m)*ak.second);
134         }
135 }
136 LL gcd(LL n,LL m)
137 {
138         if(m==0)
139                 return n;
140         else return gcd(m,n%m);
141 }
142 LL quick(LL n,LL m,LL mod)
143 {
144         LL ak = 1;
145         n %= mod;
146         while(m)
147         {
148                 if(m&1)
149                         ak =ak*n%mod;
150                 n = n*n%mod;
151                 m>>=1;
152         }
153         return ak;
154 }
155 LL er(int n,int m,LL k)
156 {
157         int mid = (n+m)/2;
158         if(n>m)return -1;
159         if(bns[mid].x == k)
160         {
161                 return bns[mid].id;
162         }
163         else if(bns[mid].x < k)
164         {
165                 return er(mid+1,m,k);
166         }
167         else return er(n,mid-1,k);
168 }
baby_step

 

posted @ 2016-10-02 18:23  sCjTyC  阅读(359)  评论(0)  编辑  收藏  举报