# 【数位DP】【P4127】[AHOI2009]同类分布

## Limitations

$1 \leq a,~b \leq 10^{18}$

## Solution

$\sum_{i = 0}^k (a_i \times 10^i \bmod B)~\equiv~0\pmod B$

$f_{i, j, k}$ 位考虑前 $i$ 位，前 $i$ 位对应模 $B$ 的值为 $j$，且后面几位的数字和为 $k$，不顶上界的方案数，转移时枚举当前这一位是几即可。

## Code

// luogu-judger-enable-o2
#include <cstdio>
#include <cstring>

const int maxn = 70;
const int maxm = 163;
const int maxt = 10;

int A[maxn], B[maxn];
ll frog[maxn][maxm][maxm];

int ReadNum(int *p);
ll calc(const int *const num, const int n);

int main() {
freopen("1.in", "r", stdin);
int x = ReadNum(A), y = ReadNum(B);
ll _sum = 0, _val = 0, _ten = 1;
for (int i = x - 1; ~i; --i) {
_sum += A[i]; _val += A[i] * _ten;
_ten *= 10;
}
qw(calc(B, y) - calc(A, x) + (!(_val % _sum)), '\n', true);
return 0;
}

int ReadNum(int *p) {
auto beg = p;
do *p = IPT::GetChar() - '0'; while ((*p < 0) || (*p > 9));
do *(++p) = IPT::GetChar() - '0'; while ((*p <= 9) && (*p >= 0));
return p - beg;
}

ll calc(const int *const num, const int n) {
int dn = n - 1;
if (n <= 1) { return num[0]; }
ll _ret = 0, _ten = 1;
for (int i = 1; i < n; ++i) _ten *= 10;
for (int p = 1; p < maxm; ++p) {
memset(frog, 0, sizeof frog);
ll ten = _ten; int tm = ten % p;
int upc = num[0] * tm % p, left = p - num[0];
for (int i = 1; i < num[0]; ++i) if (p >= i) {
frog[0][i * tm % p][p - i] = 1;
}
for (int i = 1; i < n; ++i) {
int di = i - 1;
tm = (ten /= 10) % p;
for (int j = 0; j < p; ++j) {
for (int k = 0; k < p; ++k) {
for (int h = 0; h < 10; ++h) if ((h + k) <= p) {
int dh = h * tm % p, dj = j >= dh ? j - dh : j - dh + p;
frog[i][j][k] += frog[di][dj][k + h];
}
}
}
for (int j = 1; j < 10; ++j) if (j <= p) {
++frog[i][j * tm % p][p - j];
}
for (int h = 0; h < num[i]; ++h) if (h <= left) {
int dh = h * tm % p;
++frog[i][(upc + dh) % p][left - h];
}
upc = (upc + num[i] * tm) % p; left -= num[i];
}
_ret += frog[dn][0][0];
if ((upc == 0) && (left == 0)) ++_ret;
}
return _ret;
}


## Summary

posted @ 2019-08-24 00:48  一扶苏一  阅读(146)  评论(0编辑  收藏  举报