結果

問題 No.442 和と積
ユーザー bal4u
提出日時 2019-04-17 22:50:34
言語 C
(gcc 13.3.0)
結果
WA  
(最新)
AC  
(最初)
実行時間 -
コード長 3,313 bytes
コンパイル時間 1,169 ms
コンパイル使用メモリ 32,768 KB
実行使用メモリ 6,944 KB
最終ジャッジ日時 2024-07-04 23:17:45
合計ジャッジ時間 1,681 ms
ジャッジサーバーID
(参考情報)
judge4 / judge5
このコードへのチャレンジ
(要ログイン)
ファイルパターン 結果
sample AC * 3
other AC * 17 WA * 1
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.442
// 2019.4.17 bal4u
#include <stdio.h>
#include <math.h>
#define SIZE 50
long long factor[3][SIZE];
int power[3][SIZE];
int sz[3];
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(int id, long long n)
{
int i, d, size;
int *pp;
size = 0;
if ((n & 1) == 0) {
factor[id][size] = 2;
do n >>= 1, power[id][size]++;
while ((n & 1) == 0);
size++;
}
for (pp = ptbl; n > 1 && *pp > 0; pp++) {
if (n % *pp) continue;
d = *pp;
factor[id][size] = d;
do n /= d, power[id][size]++;
while (n % d == 0);
size++;
}
if (n > 1) {
int b = (int)sqrt((double)n);
for (i = 1009; n > 1; i += 2) {
if (i > b) {
factor[id][size] = n, power[id][size++] = 1;
break;
}
if (n % i == 0) {
factor[id][size] = i;
do n /= i, power[id][size]++;
while (n % i == 0);
size++;
}
}
}
return size;
}
int f[SIZE];
int main()
{
int i, j;
long long A, B, a, b, a2, b2, ans;
scanf("%lld%lld", &A, &B);
if (A == B) {
if (A & 1) ans = A;
else ans = A << 1;
printf("%lld\n", ans);
return 0;
}
a = A, b = B;
a2 = 1; while ((a & 1) == 0) a2 <<= 1, a >>= 1;
b2 = 1; while ((b & 1) == 0) b2 <<= 1, b >>= 1;
if (a2 < b2) ans = a2;
else ans = b2;
if (ans > 1 && (a > 0x4ffffffffffffffLL || b > 0x4ffffffffffffffLL)) {
printf("%lld\n", ans); return 0;
}
sz[0] = prime_factor(0, A);
sz[1] = prime_factor(1, B);
sz[2] = prime_factor(2, A+B);
i = j = 0; while (i < sz[0] && j < sz[1]) {
while (factor[1][j] > factor[0][i] && i < sz[0]) i++;
while (factor[1][j] < factor[0][i] && j < sz[1]) j++;
if (i < sz[0] && j < sz[1] && factor[0][i] == factor[1][j]) {
power[0][i] += power[1][j], factor[1][j] = 0;
i++, j++;
}
}
ans = 1;
i = j = 0; while (i < sz[0] && j < sz[2]) {
while (factor[2][j] > factor[0][i] && i < sz[0]) i++;
while (factor[2][j] < factor[0][i] && j < sz[2]) j++;
if (i < sz[0] && j < sz[2] && factor[0][i] == factor[2][j]) {
if (power[0][i] > power[2][j]) power[0][i] = power[2][j];
while (power[0][i]--) ans *= factor[0][i];
i++, j++;
}
}
i = j = 0; while (i < sz[1] && j < sz[2]) {
while (factor[2][j] > factor[1][i] && i < sz[1]) i++;
while (factor[2][j] < factor[1][i] && j < sz[2]) j++;
if (i < sz[1] && j < sz[2] && factor[1][i] == factor[2][j]) {
if (power[1][i] > power[2][j]) power[1][i] = power[2][j];
while (power[1][i]--) ans *= factor[1][i];
}
i++, j++;
}
printf("%lld\n", ans);
return 0;
}
הההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההההה
XXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXXX
0