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用miller_rabin 和 pollard_rho对大数因式分解,再用dfs寻找答案即可。
http://poj.org/problem?id=2429
1 #include <cstdio> 2 #include <cstring> 3 #include <algorithm> 4 #include <cmath> 5 using namespace std; 6 typedef __int64 LL; 7 const int maxn = 100; 8 const int times = 10; 9 LL prime[maxn], k; 10 int cnt[maxn]; 11 LL c, d, pm, mid; 12 13 void dfs(int pos, LL t){ 14 if(pos == k){ 15 if(pm < t) pm = t; 16 return; 17 } 18 LL tem = 1; 19 int c = cnt[pos]; 20 while(c--) tem *= prime[pos]; 21 dfs(pos + 1, t); 22 if(t * tem <= mid) dfs(pos + 1, t * tem); 23 } 24 25 LL random(LL n){ 26 return (double)rand() / RAND_MAX * n + 0.5; 27 } 28 29 LL multi(LL a, LL b, LL mod){ 30 a %= mod, b %= mod; 31 LL ans = 0; 32 while(b){ 33 if(b & 1) ans += a, ans %= mod; 34 b >>= 1; 35 a <<= 1; 36 a %= mod; 37 } 38 return ans; 39 } 40 41 LL power(LL a, LL p, LL mod){ 42 a %= mod; 43 LL ans = 1; 44 while(p){ 45 if(p & 1) ans = multi(ans, a, mod), ans %= mod; 46 p >>= 1; 47 a = multi(a, a, mod); 48 a %= mod; 49 } 50 return ans; 51 } 52 53 LL gcd(LL a, LL b){ 54 if(!b) return a; 55 return gcd(b, a % b); 56 } 57 58 bool witness(LL a, LL n){ 59 LL u = n - 1; 60 while(!(u & 1)) u >>= 1; 61 LL t = power(a, u, n); 62 while(u != n - 1 && t != 1 && t != n - 1){ 63 t = multi(t, t, n); 64 u <<= 1; 65 } 66 return t == n - 1 || u & 1; 67 } 68 69 bool miller_rabin(LL n){ 70 if(n == 2) return 1; 71 if(n < 2 || !(n & 1)) return 0; 72 for(int i = 0; i < times; i++){ 73 LL p = random(n - 2) + 1; 74 if(!witness(p, n)) return 0; 75 } 76 return 1; 77 } 78 79 LL pollard_rho(LL n, LL t){ 80 LL x = random(n - 2) + 1; 81 LL y = x, i = 1, k = 2, d; 82 while(1){ 83 ++i; 84 x = (multi(x, x, n) + t) % n; 85 d = gcd(y - x, n); 86 if(1 < d && d < n) return d; 87 if(x == y) return n; 88 if(i == k){ 89 y = x; 90 k <<= 1; 91 } 92 } 93 } 94 95 void solve(){ 96 LL m = d / c; 97 LL m1 = m; 98 mid = (LL)sqrt(m); 99 k = 0; 100 memset(cnt, 0, sizeof cnt); 101 if(m % 2 == 0){ 102 prime[k++] = 2; 103 while(m % 2 == 0) m >>= 1, ++cnt[k - 1]; 104 } 105 while(!miller_rabin(m) && m > 1){ 106 int seed = 12312; 107 LL p1 = m; 108 while(p1 >= m || !miller_rabin(p1)) p1 = pollard_rho(m, seed--); 109 prime[k++] = p1; 110 while(m % p1 == 0) m /= p1, ++cnt[k - 1]; 111 } 112 if(m != 1) prime[k++] = m, ++cnt[k - 1]; 113 pm = 1; 114 dfs(0, 1); 115 LL qm = m1 / pm; 116 printf("%I64d %I64d\n", pm * c, qm * c); 117 } 118 119 int main(){ 120 //freopen("in.txt", "r", stdin); 121 //freopen("out.txt", "w", stdout); 122 while(~scanf("%I64d%I64d", &c, &d)) solve(); 123 return 0; 124 }
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原文地址:http://www.cnblogs.com/astoninfer/p/4816724.html