Divisions
Time Limit: 2000ms
Memory Limit: 262144KB
This problem will be judged on CodeForcesGym. Original ID: 100753F64-bit integer IO format: %I64d Java class name: (Any)
解题：大数质因子分解
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 100001; 5 LL mul(LL a,LL b,LL mod) { 6 if(!a) return 0; 7 return ((a&1)*b%mod + (mul(a>>1,b,mod)<<1)%mod)%mod; 8 } 9 LL quickPow(LL a,LL d,LL n) { 10 LL ret = 1; 11 while(d) { 12 if(d&1) ret = mul(ret,a,n); 13 d >>= 1; 14 a = mul(a,a,n); 15 } 16 return ret; 17 } 18 bool check(LL a,LL d,LL n) { 19 if(n == a) return true; 20 while(~d&1) d >>= 1; 21 LL t = quickPow(a,d,n); 22 while(d < n-1 && t != 1 && t != n-1) { 23 t = mul(t,t,n); 24 d <<= 1; 25 } 26 return (d&1) || t == n-1; 27 } 28 bool isP(LL n) { 29 if(n == 2) return true; 30 if(n < 2 || 0 == (n&1)) return false; 31 static int p[5] = {2,3,7,61,24251}; 32 for(int i = 0; i < 5; ++i) 33 if(!check(p[i],n-1,n)) return false; 34 return true; 35 } 36 LL gcd(LL a,LL b) { 37 if(a < 0) return gcd(-a,b);//特别注意，没这个TLE 38 return b?gcd(b,a%b):a; 39 } 40 LL Pollard_rho(LL n,LL c) { 41 LL i = 1,k = 2,x = rand()%n,y = x; 42 while(true) { 43 x = (mul(x,x,n) + c)%n; 44 LL d = gcd(y - x,n); 45 if(d != 1 && d != n) return d; 46 if(y == x) return n; 47 if(++i == k) { 48 y = x; 49 k <<= 1; 50 } 51 } 52 } 53 LL Fac[maxn],tot; 54 void factorization(LL n) { 55 if(isP(n)) { 56 Fac[tot++] = n; 57 return; 58 } 59 LL p = n; 60 while(p >= n) p = Pollard_rho(p,rand()%(n-1)+1); 61 factorization(p); 62 factorization(n/p); 63 } 64 unordered_map<LL,LL>ump; 65 int main() { 66 LL x; 67 srand(time(0)); 68 while(~scanf("%I64d",&x)){ 69 tot = 0; 70 if(x == 1) { 71 puts("1"); 72 continue; 73 } 74 if(isP(x)){ 75 puts("2"); 76 continue; 77 } 78 factorization(x); 79 ump.clear(); 80 for(int i = 0; i < tot; ++i) 81 ump[Fac[i]]++; 82 unsigned long long ret = 1; 83 for(auto &it:ump) ret *= (it.second + 1); 84 printf("%I64u ",ret); 85 } 86 return 0; 87 } 88 /* 89 999999999999999989 90 100000007700000049 91 */