結果

問題 No.575 n! / m / m / m...
ユーザー bal4u
提出日時 2019-04-20 21:57:32
言語 C
(gcc 13.3.0)
結果
AC  
実行時間 6 ms / 2,000 ms
コード長 2,471 bytes
コンパイル時間 292 ms
コンパイル使用メモリ 32,640 KB
実行使用メモリ 5,248 KB
最終ジャッジ日時 2024-10-02 06:57:58
合計ジャッジ時間 1,636 ms
ジャッジサーバーID
(参考情報)
judge5 / judge2
このコードへのチャレンジ
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ファイルパターン 結果
sample AC * 3
other AC * 23
権限があれば一括ダウンロードができます

ソースコード

diff #
プレゼンテーションモードにする

// yukicoder: No.575 n! / m / m / m...
// 2019.4.20 bal4u
#include <stdio.h>
#include <math.h>
#define SIZE 50
long long factor[SIZE]; int power[SIZE];
int ptbl[] = {
3, 5, 7, 11, 13, 17, 19, 23, 29,
31, 37, 41, 43, 47, 53, 59, 61, 67, 71,
73, 79, 83, 89, 97, 101, 103, 107, 109, 113,
127, 131, 137, 139, 149, 151, 157, 163, 167, 173,
179, 181, 191, 193, 197, 199, 211, 223, 227, 229,
233, 239, 241, 251, 257, 263, 269, 271, 277, 281,
283, 293, 307, 311, 313, 317, 331, 337, 347, 349,
353, 359, 367, 373, 379, 383, 389, 397, 401, 409,
419, 421, 431, 433, 439, 443, 449, 457, 461, 463,
467, 479, 487, 491, 499, 503, 509, 521, 523, 541,
547, 557, 563, 569, 571, 577, 587, 593, 599, 601,
607, 613, 617, 619, 631, 641, 643, 647, 653, 659,
661, 673, 677, 683, 691, 701, 709, 719, 727, 733,
739, 743, 751, 757, 761, 769, 773, 787, 797, 809,
811, 821, 823, 827, 829, 839, 853, 857, 859, 863,
877, 881, 883, 887, 907, 911, 919, 929, 937, 941,
947, 953, 967, 971, 977, 983, 991, 997, 0 };
int prime_factor(long long n)
{
int i, d, sz;
int *pp;
sz = 0;
if ((n & 1) == 0) {
factor[sz] = 2;
do n >>= 1, power[sz]++;
while ((n & 1) == 0);
sz++;
}
for (pp = ptbl; n > 1 && *pp > 0; pp++) {
if (n % *pp) continue;
d = *pp;
factor[sz] = d;
do n /= d, power[sz]++;
while (n % d == 0);
sz++;
}
if (n > 1) {
int b = (int)sqrt((double)n);
for (i = 1009; n > 1; i += 2) {
if (i > b) {
factor[sz] = n, power[sz++] = 1;
break;
}
if (n % i == 0) {
factor[sz] = i;
do n /= i, power[sz]++;
while (n % i == 0);
sz++;
}
}
}
return sz;
}
#define PI 3.1415926535897932384626433832795
#define E 2.7182818284590452353602874713526
double calc(int n, long long m, long long c)
{
int i;
double a = 0;
for (i = 2; i <= n; i++) a += log10((double)i);
return a - c*log10((double)m);
}
long long check(long long n, long long f, int p)
{
long long c = 0, d = f;
while (d <= n) c += n/d, d *= f;
return c / p;
}
int main()
{
int i, sz;
long long n, m, c, d;
double p;
scanf("%lld%lld", &n, &m);
sz = prime_factor(m);
c = 0x7fffffffffffffLL; for (i = 0; i < sz; i++) {
d = check(n, factor[i], power[i]);
if (d < c) c = d;
}
if (n <= 1000) p = calc((int)n, m, c);
else p = 0.5*log10(2*PI*n) + n*log10(n/E)
+ log10(1 + 1/(12.0*n)) - c*log10((double)m);
d = (long long)(p+1e-8);
p -= d;
p = pow(10, p);
printf("%lfe%lld\n", p, d);
return 0;
}
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