【洛谷T96628】统计

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$sumlimits_{i=1}^{n}{nchoose m} imes 9^{n-i}$

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$=2^j imes 3^k imes 5^l imes7^o$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;
}