Description
给定$n$, $m$，求十进制$n$位数每个位数之积等于k的方案数
Solution
dp+高精+数学
考虑$k=0$的情况，由于可以有若干个$0$，所以方案数为$\sum\limits_{i=1}^{n}{n\choose m}\times 9^{n-i}$
考虑另外的情况，我们将$k$分解质因数，如果$k$还有除了$2$,$3$,$5$,$7$之外的质因数那么方案数为$0$
其余的情况我们考虑一个$dp$,定义$f[i][j][k][l][o]$表示考虑前$i$位，前$i$位之积$=2^j\times 3^k\times 5^l\times7^o$的方案数是多少，显然第一维可以用滚动数组压掉，转移就类似于背包的转移即可。
由于本题十分恶心的没有取模，所以我们需要高精度计算
Code
#include <bits/stdc++.h>
// check if it is judged online
#define LOCAL
namespace shl {
const int mod = 10;
int n, k;
typedef unsigned long long ull;
ull C[55][55];
ull ans[60];
ull sum[60];
int lens = 1, lena = 1;
using std::max;
int f[32][32][32][32][50];
void add() {
lena = max(lena, lens);
for (register int i = 1; i <= lena; ++i)
ans[i] += sum[i];
for (register int i = 1; i <= lena; ++i)
ans[i + 1] += ans[i] / mod, ans[i] %= mod;
while (ans[lena + 1]) lena++;
}
void mul(ull x, ull base, int mi) {
memset(sum, 0, sizeof(sum));
lens = 1;
sum[1] = 1;
for (register int i = 1; i <= mi; ++i) {
for (register int j = 1; j <= lens; ++j) sum[j] *= base;
for (register int j = 1; j <= lens; ++j)
sum[j + 1] += sum[j] / mod, sum[j] %= mod;
while (sum[lens + 1]) lens++;
}
for (register int i = 1; i <= lens; ++i) sum[i] *= x;
for (register int i = 1; i <= lens; ++i)
sum[i + 1] += sum[i] / mod, sum[i] %= mod;
while (sum[lens + 1]) lens++;
}
void plus(int x[60], int y[60]) {
int c[100], len = 1;
memset(c, 0, sizeof(c));
len = max(x[0], y[0]);
for (register int i = 1; i <= len; ++i)
c[i] = x[i] + y[i];
for (register int i = 1; i <= len; ++i)
c[i + 1] += c[i] / mod, c[i] %= mod;
while (c[len + 1]) len++;
x[0] = len;
for (register int i = 1; i <= len; ++i)
x[i] = c[i];
}
int main() {
scanf("%d%d", &n, &k);
if (k == 0) {
for (register int i = 0; i <= n; ++i)
C[i][0] = C[i][i] = 1;
for (register int i = 1; i <= n; ++i)
for (register int j = 1; j < i; ++j)
C[i][j] = C[i - 1][j] + C[i - 1][j - 1];
for (register int i = 1; i <= n; ++i) {
mul(C[n][i], 9ll, n - i);
add();
}
for (register int i = lena; i >= 1; --i)
printf("%d", ans[i]);
puts("");
return 0;
}
int a1 = 0, a2 = 0, a3 = 0, a4 = 0;
while (k % 2 == 0) a1++, k /= 2;
while (k % 3 == 0) a2++, k /= 3;
while (k % 5 == 0) a3++, k /= 5;
while (k % 7 == 0) a4++, k /= 7;
if (k > 1) {
puts("0");
return 0;
}
f[0][0][0][0][0] = 1;
f[0][0][0][0][1] = 1;
for (register int i = 1; i <= n; ++i) {
for (register int j = a1; j >= 0; --j)
for (register int k = a2; k >= 0; --k)
for (register int l = a3; l >= 0; --l)
for (register int o = a4; o >= 0; --o) {
if (j >= 1) plus(f[j][k][l][o], f[j - 1][k][l][o]); //2
if (k >= 1) plus(f[j][k][l][o], f[j][k - 1][l][o]); //3
if (j >= 2) plus(f[j][k][l][o], f[j - 2][k][l][o]); //4
if (l >= 1) plus(f[j][k][l][o], f[j][k][l - 1][o]); //5
if (j >= 1 && k >= 1) plus(f[j][k][l][o], f[j - 1][k - 1][l][o]); //6
if (o >= 1) plus(f[j][k][l][o], f[j][k][l][o - 1]); //7
if (j >= 3) plus(f[j][k][l][o], f[j - 3][k][l][o]); //8
if (k >= 2) plus(f[j][k][l][o], f[j][k - 2][l][o]); //9
}
}
for (register int i = max(f[a1][a2][a3][a4][0], 1); i >= 1; --i)
printf("%d", f[a1][a2][a3][a4][i]);
return 0;
}
}
int main() {
#ifdef LOCAL
freopen("count.in", "r", stdin);
freopen("count.out", "w", stdout);
#endif
shl::main();
return 0;
}