Ultraviolet A 11916-emoogle grid (number theory)

Ultraviolet A 11916-emoogle Grid

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A grid with n columns can have B grids without coloring. Other grids are painted with one color. There are K colors in total, and the color of the grids in the same column cannot be the same, ask the total number of solutions mod 100000007. The answer is equal to the minimum m value when R is used.

Idea: divide the lattice into two parts: unpainted and unpainted, and then calculate the number of cases. If there are multiple rows, the color that can be painted on each grid must be k-1, that is, the total number of rows (k-1) ^ N is calculated
First part of the situation * (k-1) ^ N) ^ X % mod = R is X.
That is, the requestLOGR/The first part ((K? 1)N) %MODThe answer can be obtained by using various fast power, inverse, and discrete logarithm.

Code:

#include <stdio.h>#include <string.h>#include <math.h>#include <map>#include <set>#include <algorithm>using namespace std;long long exgcd(long long a, long long b, long long &x, long long &y) {    if (!b) {x = 1; y = 0; return a;}    long long d = exgcd(b, a % b, y, x);    y -= (a / b) * x;    return d;}long long inv(long long a, long long n) {    long long x, y;    exgcd(a, n, x, y);    return (x + n) % n;}long long pow_mod(long long x, long long k, long long n) {    if (k == 0) return 1;    long long ans = pow_mod(x * x % n, k>>1, n);    if (k&1)ans = ans * x % n;    return ans;}long long log_mod(long long a, long long b, long long n) {    long long m = (long long)sqrt(n + 0.5), v, e = 1, i;    v = inv(pow_mod(a, m, n), n);    map<long long, long long> x;    x[1] = 0;    for (long long i = 1; i < m; i++) {e = e * a % n;if (!x.count(e)) x[e] = i;    }    for (long long i = 0; i < m; i++) {if (x.count(b)) return i * m + x[b];b = b * v % n;    }    return -1;}const long long MOD = 100000007;long long t, n, k, b, r, Max, x[505], y[505];typedef pair<long long, long long> pii;set<pii> beats;long long cal() {    long long ans = 0;    for (long long i = 0; i < b; i++) {if (x[i] != Max && !beats.count(make_pair(x[i] + 1, y[i])))    ans++;    }    ans += n;    for (long long i = 0; i < b; i++) if (x[i] == 1) ans--;    return pow_mod(k, ans,  MOD) * pow_mod(k - 1, Max * n - b - ans, MOD) % MOD;}long long solve() {    long long m = cal();    if (m == r) return Max;    long long tmp = n;    for (long long i = 0; i < b; i++)if (x[i] == Max) tmp--;    long long ans = pow_mod(k - 1, tmp, MOD) * pow_mod(k, n - tmp, MOD) % MOD;    m = m * ans % MOD;    if (m == r) return Max + 1;    return log_mod(pow_mod(k - 1, n, MOD), r * inv(m, MOD) % MOD, MOD) + Max + 1;}int main() {    long long cas = 0;    scanf("%lld", &t);    while (t--) {beats.clear();Max = 1;scanf("%lld%lld%lld%lld", &n, &k, &b, &r);for (long long i = 0; i < b; i++) {    scanf("%lld%lld", &x[i], &y[i]);    beats.insert(make_pair(x[i], y[i]));    Max = max(Max, x[i]);}printf("Case %lld: %lld\n", ++cas, solve());    }    return 0;}