從1-n裏面選數,數的乘積要求是完全平方,求這個乘積的最大值,要求mod 1000000007
對n!進行素因子分解,如果素因子是奇數的話,要變成偶數,也就是n!要除以該素因子。否則的話就保留該素因子。
首先篩素數,然後對n!進行素因子計算。把要除的積保存在一個數裏面。
最後求n! / 該數,也就是對該數關於mod_number的逆,mod_number = 100000007爲素數,就相當於求a^mod - 2。
不是素數的話要進行求逆運算。
對了...被分解n!素因子坑死了。也就是getSum函數中的cnt_number,如果是long long的話會超時。。
可能long long的除法很費時吧。。
AC代碼:
#include <cstdio>
#include <string.h>
#include <cmath>
const long long MOD = 1000000007;
const int MAX_NUMBER = 10000000;
bool vis[MAX_NUMBER + 2] = {0};
int prime[MAX_NUMBER];
long long fact[MAX_NUMBER + 2];
int prime_number;
int number;
void getAllPrime() {
int m = (int)sqrt(MAX_NUMBER + 0.5);
for (int i = 2; i <= m; i++) {
if (!vis[i]) {
for (int j = i * i; j <= MAX_NUMBER; j += i) {
vis[j] = 1;
}
}
}
prime_number = 0;
for (int i = 2; i <= MAX_NUMBER; i++) {
if (!vis[i]) {
prime[++prime_number] = i;
}
}
}
long long getInverse(long long a, long long power_number) {
long long ans = 1;
while (power_number > 0) {
if (power_number & 1) {
ans = (ans * a) % MOD;
}
a = (a * a) % MOD;
power_number /= 2;
}
return ans;
}
int getSum(int cnt_number, int factor) {
int ans = 0;
while (cnt_number > 0) {
ans += cnt_number / factor;
cnt_number /= factor;
}
return ans;
}
int main() {
getAllPrime();
fact[1] = 1;
for (int i = 2; i <= MAX_NUMBER; i++) {
fact[i] = (fact[i - 1] * i) % MOD;
}
while (scanf("%d", &number) != EOF) {
if (!number) {
break;
}
long long to_dive = 1;
for (int i = 1; i <= prime_number && prime[i] <= number; i++) {
int power = getSum(number, prime[i]);
if (power & 1) {
to_dive = (to_dive * prime[i]) % MOD;
}
}
to_dive = getInverse(to_dive, MOD - 2);
printf("%I64d\n", (fact[number] * to_dive) % MOD);
}
return 0;
}