POJ 3696/ HDU 2462 The Luckiest number (数论)

题目：LINK 

给定一个数L, 求使得k*L ==8...8(一串8) 求这一串8的最小长度.

对于8.....8可以写成 (10^x -1)*8/9  

    即 (10^x - 1)*8/9 = k*n

    (10^x - 1)* 8 / gcd(8, n) = 9*n*k/gcd(8,n) ;

    令p = 8/gcd(8,n); q = 9*n/gcd(8,n); 这里p,q互质

    因此(10^x - 1) 是 q的整数倍

    可以得到 10^x = 1 (mod q) 

    只要求得最小的x(x>0) 满足上面的式子就可以了

这里有两种方法 1),利用欧拉定理: 当gcd(a,b)==1时，a^φ(b)≡1(mod b); 如果gcd(10,q) != 1 无解.否则解为φ(q) 的因子(拉格朗日定理) ,我们可以处理出来φ(q)的所有因子，依次进行判断即可.

    2),对于求解 a^x = b (mod c)，我们可以采用BSGS算法(我直接套模板) .

1) 

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<map>
using namespace std;
#define INF 1000000000
typedef __int64 LL;
#define N 2000000002
#define M 100005
LL n;
vector<LL > save, fac, allfac;
LL cou[100];
bool prime[M];
LL gcd(LL a, LL b)
{
if(!b) return a;
return gcd(b, a%b);
}
LL eular(LL n)
{
LL ret = 1;
for(LL i=2; i*i<=n; i++) {
if(n%i ==0) {
n /= i;
ret *= (i-1);
while(n%i == 0) {
n /= i; ret *= i;
}
}
}
if(n>1) ret *= (n-1) ;
return ret;
}
LL mul(LL x, LL y, LL mod)
{
LL ret = 0;
while(y ) {
if(y&1) {
ret += x; ret %= mod;
}
y >>= 1;
x += x; x %= mod;
}
return ret;
}
LL pow_(LL x, LL y, LL mod)
{
LL ret = 1;
while(y) {
if(y&1) {
ret = mul(ret, x, mod);
}
y >>= 1;
x = mul(x, x, mod);
}
return ret ;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif // ONLINE_JUDGE
int time = 0;
while(scanf("%I64d", &n), n) {
LL tmp = n*9/gcd(8, n);
printf("Case %d: ", ++time);
if(gcd(10, tmp) != 1) {
printf("0\n");
continue;
}
LL pp = eular(tmp);
allfac.clear();
for(LL i = 1; i*i <=pp; i++) {
if(pp % i == 0) {
allfac.push_back(i);
allfac.push_back(pp/i);
}
}
sort(allfac.begin(), allfac.end());
for(int i = 0; i < allfac.size(); i++) {
LL gg = pow_(10, allfac[i], tmp);
if(gg == 1) {
printf("%I64d\n", allfac[i]);
break;
}
}
}
return 0;
}

2)

#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
#include<string>
#include<vector>
#include<cmath>
#include<queue>
#include<map>
#include<set>
using namespace std;
#define INF 1000000000
typedef __int64 LL;
const LL maxn = 65535;
struct hash
{
LL a,b,next;
} Hash[maxn << 1];
LL flg[maxn + 66];
LL top,idx;
void ins(LL a,LL b)
{
LL k = b & maxn;
if(flg[k] != idx)
{
flg[k] = idx;
Hash[k].next = -1;
Hash[k].a = a;
Hash[k].b = b;
return ;
}
while(Hash[k].next != -1)
{
if(Hash[k].b == b) return ;
k = Hash[k].next;
}
Hash[k].next = ++ top;
Hash[top].next = -1;
Hash[top].a = a;
Hash[top].b = b;
}
LL find(LL b)
{
LL k = b & maxn;
if(flg[k] != idx) return -1;
while(k != -1)
{
if(Hash[k].b == b) return Hash[k].a;
k = Hash[k].next;
}
return -1;
}
LL gcd(LL a,LL b)
{
return b?gcd(b,a%b):a;
}
LL ext_gcd(LL a,LL b,LL& x,LL& y)
{
LL t,ret;
if (!b)
{
x=1,y=0;
return a;
}
ret=ext_gcd(b,a%b,x,y);
t=x,x=y,y=t-a/b*y;
return ret;
}
LL Inval(LL a,LL b,LL n)
{
LL x,y,e;
ext_gcd(a,n,x,y);
e=(LL)x*b%n;
return e<0?e+n:e;
}
LL mul(LL x, LL y, LL mod)
{
LL ret = 0;
while(y ) {
if(y&1) {
ret += x; ret %= mod;
}
y >>= 1;
x += x; x %= mod;
}
return ret;
}
LL pow_(LL x, LL y, LL mod)
{
LL ret = 1;
while(y) {
if(y&1) {
ret = mul(ret, x, mod);
}
y >>= 1;
x = mul(x, x, mod);
}
return ret ;
}
LL BabyStep(LL A,LL B,LL C)
{
top = maxn;
++ idx;
LL buf=1%C,D=buf,K;
LL i,d=0,tmp;
//buf = buf *A%C;
for(i=0; i<=100; buf=buf*A%C,++i)if(buf==B)return i;
while((tmp=gcd(A,C))!=1)
{
if(B%tmp)return -1;
++d;
C/=tmp;
B/=tmp;
D=D*A/tmp%C;
}
LL M=(LL)ceil(sqrt((double)C));
for(buf=1%C,i=0; i<=M; buf=buf*A%C,++i)ins(i,buf);
for(i=0,K=pow_((LL)A,M,C); i<=M; D=D*K%C,++i)
{
tmp=Inval((LL)D,B,C);
LL w ;
if(tmp>=0&&(w = find(tmp)) != -1)
{
if(i*M+w+d<=0) continue;
//printf("%I64d \n", i*M + w+ d);
return i*M+w+d;
}
}
return -1;
}
LL eular(LL n)
{
LL ret = 1;
for(LL i=2; i*i<=n; i++) {
if(n%i ==0) {
n /= i;
ret *= (i-1);
while(n%i == 0) {
n /= i; ret *= i;
}
}
}
if(n>1) ret *= (n-1) ;
return ret;
}
int main()
{
#ifndef ONLINE_JUDGE
freopen("in.txt", "r", stdin);
#endif // ONLINE_JUDGE
LL time = 0;
LL n;
while(scanf("%I64d", &n), n) {
LL tmp = n*9/gcd(8, n);
printf("Case %d: ", ++time);
//10^X ==1 MOD tmp
LL ans = BabyStep(10LL, 1LL, tmp);
if(ans<0) {
printf("0\n"); continue;
}
if(ans>0) {
printf("%I64d\n", ans);
}
LL P = tmp, A = 10, D = 1;
LL x = eular(P);
if(pow_(A, ans + x, P) != D) {
printf("0\n"); continue;
}
LL x1 = x;
for(LL i = 2; i * i<= x1; i++) {
if(x1 % i ==0) {
while(x1 % i == 0) x1 /= i;
while(x%i==0 && pow_(A, ans + x/i, P)==D) x/=i;
}
}
if(x1 != 1) {
while(x % x1 ==0 && pow_(A, ans + x/x1, P) ==D) x/=x1;
}
printf("%I64d\n", ans + x);
}
return 0;
}