附上蒟蒻的sb式推导过程, 如有不严谨请指出, 谢谢~
a0+a1∗pq+....an∗pnqn=0
a0=−a1∗pq...−an∗pnqn
左右同时乘以qn,a0∗qn=−a1∗p∗qn−1−...−an−1∗pn−1∗q−an∗pn
然后我们就会发现左边是q的倍数, 右边的a0到an−1也都是q的倍数, 唯独an∗pn没有q这个因子, 又因为约分后p和q互质, 那么说明an有点问题啊, 我们知道右边是q的倍数, 左边0-n-1项也都是q的倍数, 那么只有可能an∗pn也是q的倍数才能加起来是q的倍数… p跟q互质, 那么就说明an是q的倍数… 将a0和an调换位置我们就可以得到:
an∗pn=−a1∗p∗qn−1−...−an−1∗pn−1∗q−a0∗qn
同样的推理方法, 我们会发现a0是p的倍数. 那么此时p是a0的约数, q是an的约数… 暴力枚举约数就好了. 约数根号时间筛出来… 听Doggu说2e7内最多的就只有500多个约数… 那么最坏5002∗100, 稳过一颗赛艇啊… 注意a0为0的话要特判.
#include<stdio.h>
#include<cmath>
#include<algorithm>
using namespace std;
typedef long long dnt;
const int maxm = 105;
const int maxn = 1e6;
const dnt mod = 1e9 + 7;
int n, cnt, na, nb;
dnt pwx[maxm], pwy[maxm];
int a[maxn], b[maxn], c[maxm];
int gcd(int a, int b) {
return (!b) ? a : gcd(b, a % b);
}
struct query {
int u, d, f; //means up and down
friend bool operator < (const query &x, const query &y) {
if (x.f == -1 && y.f == +1)
return true;
if (x.f == +1 && y.f == -1)
return false;
if (x.f == 1 && y.f == 1)
return 1ll * x.u * y.d < 1ll * y.u * x.d;
if (x.f == -1 && y.f == -1)
return 1ll * x.u * y.d > 1ll * y.u * x.d;
}
}q[maxm];
inline void init() {
int fr = 0;
while (!c[fr]) ++ fr;
if (!fr) return;
n = n - fr;
for (int i = 0; i <= n; ++ i)
c[i] = c[i + fr];
q[++ cnt].u = 0, q[cnt].d = 1, q[cnt].f = 1;
}
inline void frac(int x, int *array, int &num) {
int lim = (int)sqrt(x);
for (int i = 1; i <= lim; ++ i) // it should be <=
if (x % i == 0) {
array[++ num] = i;
array[++ num] = x / i;
}
if (lim * lim == x) -- num;
}
inline void calc(dnt x, dnt y) {
int f = 1;
dnt ans = 0, ret = 0;
if (x < 0) f = -1, x = -x;
for (int i = 1; i <= n; ++ i)
pwx[i] = (pwx[i - 1] * x) % mod,
pwy[i] = (pwy[i - 1] * y) % mod;
for (int i = 0; i <= n; ++ i) {
ret = c[i] * pwx[i] % mod * pwy[n - i] % mod;
if (i & 1)
ret = (ret * f + mod) % mod;
ans = (ans + ret) % mod;
}
if (!ans)
q[++ cnt].u = x, q[cnt].d = y, q[cnt].f = f;
}
inline void bruce_solve() {
pwx[0] = pwy[0] = 1;
for (int i = 1; i <= na; ++ i)
for (int j = 1; j <= nb; ++ j)
if (gcd(a[i], b[j]) == 1)
calc(a[i], b[j]), calc(-a[i], b[j]);
}
int main() {
scanf("%d", &n);
for (int i = 0; i <= n; ++ i)
scanf("%d", &c[i]);
init();
frac(abs(c[0]), a, na), frac(abs(c[n]), b, nb);
bruce_solve();
sort(q + 1, q + cnt + 1);
printf("%d\n", cnt);
for (int i = 1; i <= cnt; ++ i) {
if (q[i].f == -1) putchar('-');
if (q[i].u % q[i].d == 0) printf("%d\n", q[i].u / q[i].d);
else printf("%d/%d\n", q[i].u, q[i].d);
}
return 0;
}