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【BZOJ4870】【SHOI2017】组合数问题

2018-03-23 15:12 281 查看
【题目链接】
点击打开链接

【思路要点】
原问题可等价于在\(Nk\)个物品中选出\(pk+r(p\in Z)\)个的方案数。
矩阵乘法即可,时间复杂度\(O(k^3LogN)\)。
【代码】
#include<bits/stdc++.h>
using namespace std;
const int MAXLOG = 63;
const int MAXN = 55;
template <typename T> void chkmax(T &x, T y) {x = max(x, y); }
template <typename T> void chkmin(T &x, T y) {x = min(x, y); }
template <typename T> void read(T &x) {
x = 0; int f = 1;
char c = getchar();
for (; !isdigit(c); c = getchar()) if (c == '-') f = -f;
for (; isdigit(c); c = getchar()) x = x * 10 + c - '0';
x *= f;
}
template <typename T> void write(T x) {
if (x < 0) x = -x, putchar('-');
if (x > 9) write(x / 10);
putchar(x % 10 + '0');
}
template <typename T> void writeln(T x) {
write(x);
puts("");
}
int n, P, m, r;
int vec[1][MAXN], tmp[1][MAXN];
int matrix[MAXLOG][MAXN][MAXN];
int main() {
read(n), read(P), read(m), read(r);
for (int i = 0; i < m; i++) {
matrix[0][i][i]++;
matrix[0][i][(i + 1) % m]++;
}
for (int p = 1; p < MAXLOG; p++)
for (int i = 0; i < m; i++)
for (int j = 0; j < m; j++)
for (int k = 0; k < m; k++)
matrix[p][i][j] = (matrix[p][i][j] + 1ll * matrix[p - 1][i][k] * matrix[p - 1][k][j]) % P;
long long N = 1ll * n * m;
vec[0][0] = 1;
for (int p = 0; p < MAXLOG; p++) {
long long bit = 1ll << p;
if (N & bit) {
memset(tmp, 0, sizeof(tmp));
for (int i = 0; i <= 0; i++)
for (int j = 0; j < m; j++)
for (int k = 0; k < m; k++)
tmp[i][j] = (tmp[i][j] + 1ll * vec[i][k] * matrix[p][k][j]) % P;
memcpy(vec, tmp, sizeof(tmp));
}
}
writeln(vec[0][r]);
return 0;
}
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