Hdu 4869 Turn the pokers (Mathematics), hdupokers

Hdu 4869 Turn the pokers (Mathematics), hdupokers

Link: hdu 4869 Turn the pokers

Given n and m, it indicates that there are n chances of flop turning. m cards, all cards at the beginning are facing up on the back. You can select xi cards and flip cards at the same time for each flop. How many features are there at the end.

Solution: you only need to determine the number of cards with the last positive orientation. Therefore, you can maintain the upper and lower limits when reading xi.

#include <cstdio>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;//typedef __int64 ll;const ll mod = 1e9+9;const int maxn = 1e5+10;int n, m, l, r;ll c[maxn];ll pow_mod (ll x, int k) {    ll ans = 1;    while (k) {        if (k&1)            ans = ans * x % mod;        x = x * x % mod;        k /= 2;    }    return ans;}void init () {    int x, p, q;    l = r = 0;    for (int i = 0; i < n; i++) {        scanf("%d", &x);        if (l >= x)            p = l - x;        else if (r >= x)            p = ( (l&1) == (x&1) ? 0 : 1);        else            p = x - r;        if (r + x <= m)            q = r + x;        else if (l + x <= m)            q = ( ((l+x)&1) == (m&1) ? m : m-1);        else            q = 2 * m - l - x;        l = p;        r = q;    }}ll solve () {    c[0] = 1;    ll ans = 0;    for (int i = 0; i <= r; i++) {        if (i) {            if (m - i < i)                c[i] = c[m-i];            else                //c[i] = c[i-1] * (m-i+1) % mod * pow_mod(i, mod-2)  % mod;                c[i] = c[i-1] * ( (ll)(m-i+1) * pow_mod(i, mod-2)  % mod) % mod;        }        if (i >= l && (i&1) == (l&1))            ans = (ans + c[i]) % mod;    }    return ans;}int main () {    while (scanf("%d%d", &n, &m) == 2) {        init();        printf("%lld\n", solve());    }    return 0;}