bzoj 4870: [Shoi2017]组合数问题 [矩阵乘法优化dp]
4870: [Shoi2017]组合数问题
题意:求
\[\sum_{i=0}^{n-1} \binom{nk}{ik+r} \mod p
\]
\(n \le 10^9, 0\le r < k \le 50\)
组合数推了一下,有一些有趣的性质但是并不好做
想到了从意义方面考虑,但是没有深入,去看了题解
n大k小,一副矩乘的样子
就是求“n个物品取模k余r个的方案数”
因为取的个数模k,变得很有意思,可以把组合数的递推式矩乘了...
#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long ll;
inline int read() {
char c=getchar(); int x=0,f=1;
while(c<'0' || c>'9') {if(c=='-')f=-1; c=getchar();}
while(c>='0' && c<='9') {x=x*10+c-'0'; c=getchar();}
return x*f;
}
int n, mo, k, r;
struct matrix {
int a[51][51];
matrix() {memset(a, 0, sizeof(a));}
int* operator [](int x) {return a[x];}
} f, a;
matrix operator *(matrix a, matrix b) {
int n = k;
matrix c;
for(int i=0; i<n; i++)
for(int k=0; k<n; k++) if(a[i][k])
for(int j=0; j<n; j++) if(b[k][j])
c[i][j] = (c[i][j] + (ll) a[i][k] * b[k][j] %mo) %mo;
return c;
}
matrix operator ^(matrix a, ll b) {
int n = k;
matrix c;
for(int i=0; i<n; i++) c[i][i] = 1;
for(; b; b>>=1, a=a*a) if(b&1) c=c*a;
return c;
}
int Pow(ll a, int b) {
ll ans=1;
for(; b; b>>=1, a=a*a%mo)
if(b&1) ans=ans*a%mo;
return ans;
}
int main() {
freopen("in", "r", stdin);
n=read(); mo=read(); k=read(); r=read();
if(k == 1) {printf("%d", Pow(2, n)); return 0;}
for(int i=0; i<k; i++) f[i][i] = 1, f[(i+1)%k][i] = 1;
f = f ^ ((ll) n * k);
a[0][0] = 1;
a = f * a;
printf("%d\n", a[r][0]);
}
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