Hdu 4832 Chess (count + dp)

Link: hdu 4832 Chess

Question: omitted. (Note that King can only go around 8 cells)

Solution: Separate the horizontal and vertical values. l [I] indicates the number of types that go vertical without going out of bounds. r [I] indicates the number of types that go horizontally in step I without going out of bounds, the number of horizontal and vertical steps (k) are enumerated and multiplied by the combination number. Because the number of steps is determined, but the relationship between steps must be considered.
When processing the number of steps, open a two-dimensional array dp [I] [j], indicating the number of steps in the position of j, j is the offset, if j-k is negative, it indicates the value is on the left/of the start point. | j? K | position, where j-k is a positive number, indicating the right/Bottom of the starting point | j? K |.

#include 
  
   #include 
   
    #include 
    
     using namespace std;typedef long long ll;const int N = 1010;const ll MOD = 9999991;int n, m, k, x, y;ll l[N], r[N], g[N][N*2], c[N][N];void cat (ll* a, int x, int t) {    memset(g, 0, sizeof(g));    g[0][k] = 1;    int up = k-(x-1);    int down = k+(t-x);    for (int i = 1; i <= k; i++) {        for (int j = up; j <= down; j++) {            if (g[i-1][j]) {                if (j != up) {                    g[i][j-1] = (g[i][j-1] + g[i-1][j]) % MOD;                    if (j != up+1)                        g[i][j-2] = (g[i][j-2] + g[i-1][j]) % MOD;                }                if (j != down) {                    g[i][j+1] = (g[i][j+1] + g[i-1][j]) % MOD;                    if (j != down-1)                        g[i][j+2] = (g[i][j+2] + g[i-1][j]) % MOD;                }            }            a[i-1] = (a[i-1] + g[i-1][j]) % MOD;        }    }    for (int i = up; i <= down; i++)        a[k] = (a[k] + g[k][i]) % MOD;}void input () {    for (int i = 0; i < N; i++) {        c[i][0] = c[i][i] = 1;        for (int j = 0; j < i; j++)            c[i][j] = (c[i-1][j-1] + c[i-1][j]) % MOD;    }}void init () {    //scanf("%d%d%d%d%d", &n, &m, &k, &x, &y);    cin >> n >> m >> k >> x >> y;    memset(l, 0, sizeof(l));    memset(r, 0, sizeof(r));    cat(l, x, n);    cat(r, y, m);}ll solve () {    ll ans = 0;    for (int i = 0; i <= k; i++)        ans = (ans + (l[i] * r[k-i]) % MOD * c[k][i])%MOD;    return ans;}int main () {    input();    int cas;    cin >> cas;    for (int i = 1; i <= cas; i++) {        init();        //printf("Case #%d:\n%lld\n", i, solve());        cout << "Case #" << i << ":" << endl;        cout << solve() << endl;    }    return 0;}