Bzoj1300 [llh Invitational competition] Big Data Calculator

At first, I thought it was a brute force attack. Later I saw the data range and thought it was necessary to implement FFT. I was abused by various types of attacks. Then, the orz seter was a Great God !!!

I only want the first three digits: a * B <=> 10 ^ (log (A) + Log (B )), so we can log all the multiplication numbers first and then change them back...

Then the last nine digits are used:

Seter: "For Prime Number A, in N! N/A + N/A ^ 2 +... times"

So C (m, n) = n! /M! /(N-m )! You can find the number of occurrences of each prime factor!

 

 1 /************************************************************** 2     Problem: 1300 3     User: rausen 4     Language: C++ 5     Result: Accepted 6     Time:164 ms 7     Memory:1792 kb 8 ****************************************************************/ 9  10 #include <cstdio>11 #include <cmath>12 #include <algorithm>13  14 using namespace std;15 typedef long long LL;16  17 int n, m;18 LL ans = 1;19 bool p[1000001];20  21 int main(){22     scanf("%d%d", &n, &m);23     m = min(m, n - m);24      25     int i, j, x, y, c, s;26     double k = 0, f = 1;27     for (i = 2; i <= n; ++i)28         if (!p[i]){29             for (j = n, x = n - m, y = m; j;){30                 c = (j /= i) - (x /= i) - (y /= i), s = i;31                 for (c <<= 1; c >>= 1; s *= s)32                     if (c & 1) ans *= s, ans %= (LL) 1e12;33             }34             if (i <= 1000)35                 for (j = i * i; j <= n; j += i)36                     p[j] = 1;37         }38     for (i = 1; i <= m; ++i){39         if (f > 1e7)40             k += log10(f), f = 1;41         f = f * (n - m + i) / i;42     }43     k += log10(f);44      45     if (k < 13) printf("%lld\n", ans);46     else printf("%d...%09lld\n", (int) (pow(10, k - floor(k) + 2) + 1e-5), ans % (int) 1e9);47     return 0;48 }

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P.s. The seter can be used in both C and C ++...

Bzoj1300 [llh Invitational competition] Big Data Calculator