bzoj 3398

f[i]表示最后一个是公牛的方案数，=sigma(f[j])(j<i-k) 然后前缀和优化即可。

#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <algorithm>
#include <cmath>

#define Mod 5000011

using namespace std;
const int N = 1e5 + 10;

int f[N], sum[N];

int main() {
    int n, K;
    scanf("%d%d", &n, &K);
    K ++;
    f[0] = sum[0] = 1;
    for(int i = 1; i <= n; i ++) {
        int j = i - K;
        if(j < 0) f[i] = 1;
        else f[i] = sum[j];
        sum[i] = (f[i] + sum[i - 1]) % Mod;
    }
    printf("%d", sum[n]);
    return 0;
}