标签:ons linker chm called clu its title names --
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
题意:一个数的因数是否为4个;是,输出>1的因数;
思路:Pollard_rho算法和Miller_Rabin算法,前一个大质数分解,后面判断大数是否为质数;
acdream的模板;
#include<iostream> #include<cstdio> #pragma comment(linker, "/STACK:1024000000,1024000000") #include<iostream> #include<cstdio> #include<cmath> #include<string> #include<queue> #include<algorithm> #include<stack> #include<cstring> #include<vector> #include<list> #include<set> #include<map> #include<bitset> using namespace std; #define LL unsigned long long #define pi (4*atan(1.0)) #define eps 1e-4 #define bug(x) cout<<"bug"<<x<<endl; const int N=5500,M=1e5+10,inf=1e9+10; const LL INF=1e18+10,mod=2147493647; const int Times = 10; LL ct, cnt; LL fac[N], num[N]; LL gcd(LL a, LL b) { return b? gcd(b, a % b) : a; } LL multi(LL a, LL b, LL m) { LL ans = 0; a %= m; while(b) { if(b & 1) { ans = (ans + a) % m; b--; } b >>= 1; a = (a + a) % m; } return ans; } LL quick_mod(LL a, LL b, LL m) { LL ans = 1; a %= m; while(b) { if(b & 1) { ans = multi(ans, a, m); b--; } b >>= 1; a = multi(a, a, m); } return ans; } bool Miller_Rabin(LL n) { if(n == 2) return true; if(n < 2 || !(n & 1)) return false; LL m = n - 1; int k = 0; while((m & 1) == 0) { k++; m >>= 1; } for(int i=0; i<Times; i++) { LL a = rand() % (n - 1) + 1; LL x = quick_mod(a, m, n); LL y = 0; for(int j=0; j<k; j++) { y = multi(x, x, n); if(y == 1 && x != 1 && x != n - 1) return false; x = y; } if(y != 1) return false; } return true; } LL pollard_rho(LL n, LL c) { LL i = 1, k = 2; LL x = rand() % (n - 1) + 1; LL y = x; while(true) { i++; x = (multi(x, x, n) + c) % n; LL d = gcd((y - x + n) % n, n); if(1 < d && d < n) return d; if(y == x) return n; if(i == k) { y = x; k <<= 1; } } } void Find(LL n, int c) { if(n == 1) return; if(Miller_Rabin(n)) { fac[ct++] = n; return ; } LL p = n; LL k = c; while(p >= n) p = pollard_rho(p, c--); Find(p, k); Find(n / p, k); } int main() { LL n; while(~scanf("%I64d",&n)) { ct = 0; Find(n, 120); sort(fac, fac + ct); num[0] = 1; int k = 1; for(int i=1; i<ct; i++) { if(fac[i] == fac[i-1]) ++num[k-1]; else { num[k] = 1; fac[k++] = fac[i]; } } cnt = k; LL ans = 0; if(cnt==1) { if(num[0]==3)printf("%lld %lld %lld\n",fac[0],fac[0]*fac[0],n); else printf("is not a D_num\n"); } else if(cnt==2) { if(num[0]==1&&num[1]==1)printf("%lld %lld %lld\n",fac[0],fac[1],n); else printf("is not a D_num\n"); } else printf("is not a D_num\n"); //for(int i=0;i<cnt;i++) //cout<<num[i]<<" "<<fac[i]<<endl; } return 0; }
hdu 3864 D_num Pollard_rho算法和Miller_Rabin算法
标签:ons linker chm called clu its title names --
原文地址:http://www.cnblogs.com/jhz033/p/7107047.html