Ultraviolet A 11916-emoogle grid (STRIDE small step algorithm)

Question connection: Ultraviolet A 11916-emoogle Grid

One problem is that K colors are applied to the grid of n columns in the m row, where B grids do not need to be colored, and each other lattice is colored, two Adjacent grids of the same column cannot be painted with the same color. The positions of m, n, k, and B grids are given, and the result R of the overall solution is calculated based on the mathematical model of 1e8 + 7. Now we know R, and find the smallest M.

Solution: Make sure that the area without colored grids is used as the unchanged part. The total number is calculated as TMP, and the first line of the variable part is added. The number of solutions is CNT. The variable part is not included in the first line, the total number of rows added is multiplied by (K? 1) n, existing

• CNT? PM = rmod (1e8 + 7)
• PM = CNT? 1? Rmod (1e8 + 7)
  That is, the stride small step algorithm is used to calculate M.

#include <cstdio>#include <cstring>#include <cmath>#include <set>#include <map>#include <algorithm>using namespace std;typedef long long ll;const ll MOD = 1e8+7;const int maxn = 505;int N, M, K, R, B, X[maxn], Y[maxn];set<pair<int, int> > bset;inline ll mul_mod (ll a, ll b) {    return (ll)a * b % MOD;}inline ll pow_mod (ll a, ll n) {    ll ans = 1;    while (n) {        if (n&1)            ans = mul_mod(ans, a);        a = mul_mod(a, a);        n /= 2;    }    return ans;}void gcd (ll a, ll b, ll& d, ll& x, ll& y) {    if (!b) {        d = a;        x = 1;        y = 0;    } else {        gcd (b, a%b, d, y, x);        y -= x * (a/b);    }}inline ll inv (ll a) {    ll d, x, y;    gcd(a, MOD, d, x, y);    return d == 1 ? (x+MOD)%MOD : -1;}void init () {    scanf("%d%d%d%d", &N, &K, &B, &R);    bset.clear();    M = 1;    for (int i = 0; i < B; i++) {        scanf("%d%d", &X[i], &Y[i]);        M = max(M, X[i]);        bset.insert(make_pair(X[i], Y[i]));    }}int count () {    int c = 0;    for (int i = 0; i < B; i++) {        if (X[i] != M && !bset.count(make_pair(X[i]+1, Y[i])))            c++;    }    c += N;    for (int i = 0; i < B; i++)        if (X[i] == 1)            c--;    return mul_mod(pow_mod(K, c), pow_mod(K-1, (ll)M*N-B-c));}int log_mod (ll a, ll b) {    ll m = (ll)sqrt(MOD+0.5), v, e = 1;    v = inv(pow_mod(a, m));    map<ll, ll> g;    g[1] = 0;    for (ll i = 1; i < m; i++) {        e = mul_mod(e, a);        if (!g.count(e))            g[e] = i;    }    for (ll i = 0; i < m; i++) {        if (g.count(b))            return i*m+g[b];        b = mul_mod(b, v);    }    return -1;}int doit () {    int cnt = count();    if (cnt == R)        return M;    int c = 0;    for (int i = 0; i < B; i++)        if (X[i] == M)            c++;    M++;    cnt = mul_mod(cnt, pow_mod(K, c));    cnt = mul_mod(cnt, pow_mod(K-1, N-c));    if (cnt == R)        return M;    return log_mod(pow_mod(K-1, N), mul_mod(R, inv(cnt))) + M;}int main () {    int cas;    scanf("%d", &cas);    for (int i = 1; i <= cas; i++) {        init();        printf("Case %d: %d\n", i, doit());    }    return 0;}