首页 > 代码库 > poj 2417
poj 2417
1 Accepted 8508K 391MS C++ 2004B 2 相比下边,,优化太多太多了。。。 3 /** 4 baby-step-giant-step 因为数据量太大,,自己写hash 5 6 **/ 7 #include <iostream> 8 #include <cstdio> 9 #include <cmath> 10 #include <cstring> 11 using namespace std; 12 long long n,a,b; 13 const int maxn = 499991; 14 bool Hash[maxn]; 15 long long idx[maxn]; 16 long long val[maxn]; 17 18 void ex_gcd(long long a,long long b,long long &x,long long &y){ 19 if(b==0){ 20 x=1; 21 y=0; 22 return ; 23 } 24 ex_gcd(b,a%b,x,y); 25 long long tmp = x-(a/b)*y; 26 x = y; 27 y = tmp; 28 } 29 30 long long euler(long long n){ 31 long long i,tmp = n; 32 for(i=2;i*i<=n;i++)if(n%i==0){ 33 tmp = tmp/i*(i-1); 34 while(n%i==0) 35 n = n/i; 36 } 37 if(n>1) 38 tmp = tmp/n*(n-1); 39 return tmp; 40 } 41 42 void Insert(long long id,long long num){ 43 long long k = num%maxn; 44 while(Hash[k]&&val[k]!=num){ 45 k++; 46 if(k==maxn) k = k-maxn; 47 } 48 if(!Hash[k]){ 49 Hash[k] =1; 50 idx[k] = id; 51 val[k] = num; 52 } 53 } 54 55 long long found(long long num){ 56 long long k = num%maxn; 57 while(Hash[k]&&val[k]!=num){ 58 k++; 59 if(k==maxn) k = k-maxn; 60 } 61 if(!Hash[k]){ 62 return -1; 63 } 64 return idx[k]; 65 } 66 67 long long baby_step(long long a,long long b,long long n){ 68 long long m = ceil(sqrt(euler(n)+0.5)); 69 memset(Hash,false,sizeof(Hash)); 70 memset(idx,-1,sizeof(idx)); 71 memset(val,-1,sizeof(val)); 72 long long d=1; 73 for(long long i=0;i<m;i++){ 74 Insert(i,d); 75 d = d*a%n; 76 } 77 long long res =1; 78 long long x,y; 79 for(long long i=0;i<m;i++){ 80 ex_gcd(res,n,x,y); 81 long long tmp = x*b%n; 82 tmp = (tmp%n+n)%n; 83 long long k = found(tmp); 84 if(k!=-1){ 85 return (i)*m+k; 86 } 87 res = res*d%n; 88 } 89 return -1; 90 } 91 92 int main() 93 { 94 while(scanf("%I64d%I64d%I64d",&n,&a,&b)==3){ 95 long long res = baby_step(a,b,n); 96 if(res==-1) 97 printf("no solution\n"); 98 else 99 printf("%I64d\n",res); 100 } 101 return 0; 102 } 103 104 -----------------------------------分割线--------------------------------------- 105 /** 106 高次方程。。。baby-step-giant-step 算法 107 Accepted 4592K4 516MS C++1104B 108 **/ 109 #include <iostream> 110 #include <cstdio> 111 #include <math.h> 112 #include <map> 113 using namespace std; 114 115 long long powmod(long long a,long long b,long long n){ 116 if(b==0) 117 return 1; 118 long long c =1; 119 while(b){ 120 if(b&1) 121 c =c*a%n; 122 a =a*a%n; 123 b>>=1; 124 } 125 return c; 126 } 127 128 long long logmod(long long a,long long b,long long n){ 129 long long m,v,e=1,i; 130 m = ceil(sqrt(n+0.5)); 131 //cout<<(double)(n-1)*1.0/m<<endl; 132 //long long m_n = powmod(a,m,n); 133 v = powmod(a,n-1-m,n); 134 map<long long ,long long >x; 135 x.clear(); 136 x[1] =m; 137 for(i=1;i<m;i++){ 138 e = e*a%n; 139 if(!x[e]) x[e]=i; 140 } 141 for(i=0;i<m;i++){ 142 if(x[b]){ 143 long long num = x[b]; 144 x.clear(); 145 return i*m+(m==num?0:num); 146 } 147 b = b*v%n; 148 } 149 return -1; 150 } 151 152 int main() 153 { 154 long long a,b,n; 155 while(scanf("%I64d%I64d%I64d",&n,&a,&b)==3){ 156 long long res = logmod(a,b,n); 157 if(res==-1) 158 printf("no solution\n"); 159 else 160 printf("%I64d\n",res); 161 } 162 return 0; 163 }
声明:以上内容来自用户投稿及互联网公开渠道收集整理发布,本网站不拥有所有权,未作人工编辑处理,也不承担相关法律责任,若内容有误或涉及侵权可进行投诉: 投诉/举报 工作人员会在5个工作日内联系你,一经查实,本站将立刻删除涉嫌侵权内容。