2008年第33届ACM/ICPC亚洲区预赛(合肥)网络预选赛_1007 The Luckies number

BT数论题

求解10^x = 1 (mod 9*L/gcd(L,8))的满足x>0的最小解就是答案

由8构成的数A设有x位
那么A=8(10^0+10^1+...+10^(x-1));
很容易得到A=(8/9)*(10^x-1);
题目的要求就是A=0(mod L)
就是(8/9)*(10^x-1)=0(mod L);
->8*(10^x-1)=0(mod 9L);
->10^x-1=0(mod 9L/gcd(L,8));
->10^x =1 (mod 9L/gcd(L,8)); 


解法是先求phi(9L/gcd(L,8)),然后枚举其因子

比如:L=11;9L/gcd(L,8)=99;phi(9L/gcd(L,8))=10;解得10^2 =1 (mod 99),2是满足条件的最小的因子了

顺便给出1999999999的答案:
161290320

其间要求 素数选取 求欧拉函数 素因子分解 解模方程...涉及很多方面的综合题...

 

1. #include <iostream>
2. using namespace std;
3. typedef unsigned long long llong;
4. bool prime[45000]={true,true};
5. llong l,fac[32][2],ans;
6. double eul;
7. int p[45000],top,len=0;
8. llong pro(llong x,llong y,llong n)
9. {
10.     llong ret=0,tmp=x%n;
11.     for(;y;y>>=1)
12.         {
13.             if(y&0x1)
14.                 if((ret+=tmp)>n)ret-=n;
15.             if((tmp<<=1)>n)tmp-=n;
16.         }
17.     return ret;
18. }
19. llong pow_mod(llong a,llong b,llong c)
20. {
21.     llong ret;
22.     for(ret=1;b;b>>=1)
23.     {if(b&1) ret=pro(ret,a,c);a=pro(a,a,c);}
24.     return ret;
25. }
26. llong gcd(llong a,llong b){return b?gcd(b,a%b):a;}
27. void dfs(int dep,llong val)
28. {
29.     int i;llong ret=1;
30.     if(dep==top)
31.         {
32.             if(pow_mod(10,val,l)==1&&val<ans)ans=val;
33.             return;
34.         }
35.     for(i=0;i<=fac[dep][0];i++)
36.         {
37.             dfs(dep+1,val*ret);    
38.             ret*=fac[dep][1];
39.         }
40. }
41. int main()
42. {
43.     llong n;
44.     int i,j,id=0;
45.     for(i=2;i<=213;i++)if(!prime[i])for(j=i;j*i<45000;j++)prime[i*j]=true;
46.     for(i=2;i<45000;i++)if(!prime[i])p[len++]=i;
47.     while(scanf("%lld",&n)!=EOF,n)
48.         {
49.             printf("Case %d: ",++id);
50.             ans=10000000000;
51.             top=0;
52.             l=9*n/gcd(n,8);
53.             eul=n=l;
54.             for(i=0;i<len;i++)
55.                 {
56.                     if(n%p[i]==0)
57.                         {
58.                             fac[top][0]=0;
59.                             fac[top][1]=p[i];
60.                             while(n%p[i]==0)
61.                                 {
62.                                     fac[top][0]++;
63.                                     n/=p[i];
64.                                 }
65.                             top++;
66.                             if(n==1||(n<45000&&!prime[n]))break;
67.                         }
68.                 }
69.             if(n!=1){fac[top][0]=1;fac[top][1]=n;top++;}
70.             for(i=0;i<top;i++)eul*=(1.0-1.0/(double)fac[i][1]);
71.             n=eul;
72.             top=0;
73.             for(i=0;i<len;i++)
74.                 {
75.                     if(n%p[i]==0)
76.                         {
77.                             fac[top][0]=0;
78.                             fac[top][1]=p[i];
79.                             while(n%p[i]==0)
80.                                 {
81.                                     fac[top][0]++;
82.                                     n/=p[i];
83.                                 }
84.                             top++;
85.                             if(n==1||(n<45000&&!prime[n]))break;
86.                         }
87.                 }
88.             if(n!=1){fac[top][0]=1;fac[top][1]=n;top++;}
89.             dfs(0,1);
90.             printf("%lld/n",ans==10000000000?0:ans);
91.         }
92. }