https://ac.nowcoder.com/acm/contest/885/C
这个跟平时那种求离散对数的bsgs不一样,虽然可以转化成离散对数来做不过会T掉。展开递推式然后合并具有a的项,发现其实是离散对数。
#include<cstdio>
#include<algorithm>
const int AA = 1000000;
std::pair<int, int> d[AA];
int mypow(long long x, int y, int p) {
long long res = 1;
while(y) {
if(y & 1)
res = res * x % p;
x = x * x % p;
y >>= 1;
}
return res;
}
int inv(int x, int p) {
return mypow(x, p - 2, p);
}
int val[AA], pos[AA];
void solve() {
long long n, x0, a, b, p;
int Q;
scanf("%lld%lld%lld%lld%lld%d", &n, &x0, &a, &b, &p, &Q);
if(a == 0) {
while(Q--) {
int v;
scanf("%d", &v);
if(v == x0)
puts("0");
else if(v == b)
puts("1");
else
puts("-1");
}
return;
}
long long now = x0;
int m = std::min((long long)AA, n);
for(int i = 0; i < m; i++) {
d[i] = {now, i};
now = (now * a + b) % p;
}
sort(d, d + m);
{
int new_m = 0;
for(int i = 0; i < m; i++) {
val[new_m] = d[i].first;
pos[new_m++] = d[i].second;
while(i + 1 < AA && d[i + 1].first == d[i].first)
i++;
}
m = new_m;
}
int BB = p / AA + 3;
int inv_a = inv(a, p);
int inv_b = (p - b) % p * inv_a % p;
long long aa = 1, bb = 0;
for(int i = 0; i < AA; i++) {
aa = aa * inv_a % p;
bb = (bb * inv_a + inv_b) % p;
}
while(Q--) {
int v;
scanf("%d", &v);
int it = std::lower_bound(val, val + m, v) - val;
if(it < m && val[it] == v) {
printf("%d\n", pos[it]);
continue;
}
if(n < AA) {
puts("-1");
continue;
}
bool suc = false;
for(int i = 1; i <= BB; i++) {
v = (aa * v + bb) % p;
it = std::lower_bound(val, val + m, v) - val;
if(it < m && val[it] == v) {
suc = true;
int res = i * AA + pos[it];
if(res >= n)
res = -1;
printf("%d\n", res);
break;
}
}
if(!suc)
puts("-1");
}
}
int main() {
int T;
scanf("%d", &T);
while(T--)
solve();
return 0;
}
但这里要学的不是套这个模板,而是用BSGS算法的思路去改。