标签：des style blog http color io os ar java
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Description has only two SentencesTime Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)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
Source
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构造等比数列，先设p=y/(x-1);那么有(An+p)=x(An-1+p) 即 An+p=x^n(a0+p)
若An≡0（mod a0) 即 a0*x^n+p*(x^n - 1) ≡ 0 ≡ p*(x^n - 1)(mod a0)
同余方程两边同除gcds=gcd(p,a0),可得 p/gcds * (x^n - 1) ≡ 0 (mod a0/gcds) *
∵ (p/gcds,a0/gcds) = 1
∴ *式可化为 x^n-1≡0 (mod ap=a0/gcds) 即 x^n ≡ 1 (mod ap)
根据欧拉定理可知 n = phi(ap) ，下面同反证法证明所求的k满足 k|n,k<=n
假设求得最小的满足题意的k 且 k \ n （记\为不整除）
那么一定存在t使得k*t恰大于n，那么k*t-n也一定满足题意，但kt-n<k与条件矛盾，即k|n，k<=n
1 #include<set> 2 #include<cmath> 3 #include<queue> 4 #include<cstdio> 5 #include<cstdlib> 6 #include<cstring> 7 #include<iostream> 8 #include<algorithm> 9 using namespace std; 10 typedef long long lld; 11 #define For(i,n) for(int i=1;i<=n;i++) 12 #define Rep(i,l,r) for(int i=l;i<=r;i++) 13 lld gcd(lld a,lld b){ 14 if(!b) return a; 15 else return gcd(b,a%b); 16 } 17 18 lld Min(lld a,lld b){return (a<b)?(a):(b);} 19 20 lld phi(lld n){ 21 lld ans=n;lld lim=sqrt(n); 22 Rep(i,2,lim){ 23 if(n%i==0) { 24 ans=ans*(i-1)/i; 25 while(n%i==0) n/=i; 26 } 27 } 28 if(n>1) ans=ans*(n-1)/n; 29 return ans; 30 } 31 32 lld x,y,a0,ap; 33 34 lld qpow(lld a,lld b){ 35 lld ans=1; 36 while(b){ 37 if(b&1) ans=(ans*a)%ap; 38 a=a*a%ap; 39 b>>=1; 40 } 41 return ans%ap; 42 } 43 44 int main(){ 45 while(~scanf("%I64d%I64d%I64d",&x,&y,&a0)){ 46 lld p=y/(x-1); 47 lld gcds=gcd(a0,p); 48 ap=a0/gcds; 49 if(gcd(x,ap)!=1){ 50 puts("Impossible!"); 51 continue; 52 } 53 lld ans=phi(ap);lld Lim=ans; 54 Rep(i,2,sqrt(Lim)) 55 if(Lim%i==0){ 56 if(qpow(x,i)==1) ans=Min(ans,i); 57 if(qpow(x,Lim/i)==1) ans=Min(ans,Lim/i); 58 } 59 printf("%I64d\n",ans); 60 } 61 return 0; 62 }
HDU3307 Description has only two Sentences 欧拉函数 欧拉定理
标签：des style blog http color io os ar java
原文地址：http://www.cnblogs.com/kingnight/p/4035425.html
