题目链接

BZOJ4870

题解

\[ans = \sum\limits_{i = 0}^{\infty}{nk \choose ik + r} \pmod p\]
发现实际是求
\[ans = \sum\limits_{i = 0}^{\infty}{nk \choose i}[i \mod k = r] \pmod p\]
设\(f[i][j]\)表示\(i\)个数选出\(x \mod k = j\)个数的方案数
利用组合数递推 + 矩乘转移即可

#include<algorithm>
#include<iostream>
#include<cstdlib>
#include<cstring>
#include<cstdio>
#include<vector>
#include<queue>
#include<cmath>
#include<map>
#define LL long long int
#define REP(i,n) for (int i = 1; i <= (n); i++)
#define Redge(u) for (int k = h[u],to; k; k = ed[k].nxt)
#define cls(s,v) memset(s,v,sizeof(s))
#define mp(a,b) make_pair<int,int>(a,b)
#define cp pair<int,int>
using namespace std;
const int maxn = 55,maxm = 100005,INF = 0x3f3f3f3f;
inline int read(){
int out = 0,flag = 1; char c = getchar();
while (c < 48 || c > 57){if (c == '-') flag = 0; c = getchar();}
while (c >= 48 && c <= 57){out = (out << 1) + (out << 3) + c - 48; c = getchar();}
return flag ? out : -out;
}
int n,r,K,P;
struct Matrix{
int s[maxn][maxn],n,m;
Matrix(){cls(s,0);n = m = 0;}
}A,F0,F;
inline Matrix operator *(const Matrix& a,const Matrix& b){
Matrix c;
if (a.m != b.n) return c;
c.n = a.n; c.m = b.m;
for (int i = 0; i < c.n; i++)
for (int j = 0; j < c.m; j++)
for (int k = 0; k < a.m; k++)
c.s[i][j] = (c.s[i][j] + 1ll * a.s[i][k] * b.s[k][j] % P) % P;
return c;
}
inline Matrix qpow(Matrix a,LL b){
Matrix re; re.n = re.m = a.n;
for (int i = 0; i < re.n; i++) re.s[i][i] = 1;
for (; b; b >>= 1,a = a * a)
if (b & 1) re = re * a;
return re;
}
int main(){
n = read(); P = read(); K = read(); r = read();
F0.n = K; F0.m = 1; F0.s[0][0] = 1;
A.n = A.m = K;
for (int j = 0; j < K; j++){
A.s[j][j]++;
A.s[j][(j - 1 + K) % K]++;
}
F = qpow(A,1ll * n * K) * F0;
printf("%d\n",F.s[r][0]);
return 0;
}