Description has only two Sentences
Time Limit: 3000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 852 Accepted Submission(s): 259
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
HDOJ Monthly Contest – 2010.02.06
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因为考试放下了挺久，后来发现不做题好空虚寂寞...于是决定在做一段时间再说。
1 //31MS 236K 1482 B G++
2 /*
3
4 又是不太懂的数学题，求ak，令ak%a0==0
5
6 欧拉函数+快速幂：
7 欧拉函数相信都知道了。
8 首先这题推出来的公式为：
9 ak=a0+y/(x-1)*(x^k-1);
10
11 明显a0是可以忽略的，其实就是要令
12 y/(x-1)*(x^k-1) % a0 == 0;
13 可令 m=a0/(gcd(y/(x-1),a0))，然后就求k使得
14 (x^k-1)%m==0 即可
15 即 x^k==1(mod m)
16
17 又欧拉定理有：
18 x^euler(m)==1(mod m) (x与m互质，不互质即无解)
19
20 由抽屉原理可知 x^k 的余数必在 euler(m) 的某个循环节循环。
21 故求出最小的因子k使得 x^k%m==1，即为答案
22
23 */
24 #include<stdio.h>
25 #include<stdlib.h>
26 #include<math.h>
27 __int64 e[1005],id;
28 int cmp(const void*a,const void*b)
29 {
30 return *(int*)a-*(int*)b;
31 }
32 __int64 euler(__int64 n)
33 {
34 __int64 m=(__int64)sqrt(n+0.5);
35 __int64 ret=1;
36 for(__int64 i=2;i<=m;i++){
37 if(n%i==0){
38 ret*=i-1;n/=i;
39 }
40 while(n%i==0){
41 ret*=i;n/=i;
42 }
43 }
44 if(n>1) ret*=n-1;
45 return ret;
46 }
47 __int64 Gcd(__int64 a,__int64 b)
48 {
49 return b==0?a:Gcd(b,a%b);
50 }
51 void find(__int64 n)
52 {
53 __int64 m=(__int64)sqrt(n+0.5);
54 id=0;
55 for(__int64 i=1;i<m;i++)
56 if(n%i==0){
57 e[id++]=i;
58 e[id++]=n/i;
59 }
60 if(m*m==n) e[id++]=m;
61 }
62 __int64 Pow(__int64 a,__int64 b,__int64 mod)
63 {
64 __int64 t=1;
65 while(b){
66 if(b&1) t=(t*a)%mod;
67 a=(a*a)%mod;
68 b/=2;
69 }
70 return t;
71 }
72 int main(void)
73 {
74 __int64 x,y,a;
75 while(scanf("%I64d%I64d%I64d",&x,&y,&a)!=EOF)
76 {
77 if(y==0){
78 puts("1");
79 continue;
80 }
81 __int64 m=a/(Gcd(y/(x-1),a));
82 if(Gcd(m,x)!=1){
83 puts("Impossible!");
84 continue;
85 }
86 __int64 p=euler(m);
87 find(p);
88 qsort(e,id,sizeof(e[0]),cmp);
89 for(int i=0;i<id;i++){
90 if(Pow(x,e[i],m)==1){
91 printf("%I64d\n",e[i]);
92 break;
93 }
94 }
95 }
96 return 0;
97 }
hdu 3307 Description has only two Sentences (欧拉函数+快速幂),布布扣,bubuko.com