Atcoder681 Typical DP Contest E.数 数位dp

写什么递归....非递归多好写

令$f[i][j]$表示前$i$位的和在模$d$意义下为$j$的方案数，然后转移即可

复杂度$O(10000 * 100 * 10)$

注意非递归建议高位摆第$n$位...

#include <cstdio>
#include <cstring>
using namespace std;

#define ri register int
#define rep(io, st, ed) for(ri io = st; io <= ed; io ++)
#define drep(io, ed, st) for(ri io = ed; io >= st; io --)

const int sid = 10050;
const int pid = 105;
const int mod = 1e9 + 7;

char s[sid];
int d, n, f[sid][pid];

inline void inc(int &a, int b) { a += b; if(a >= mod) a -= mod; }
inline int inv(int a) { return (d - (a % d)) % d; }

void Solve() {
    f[0][0] = 1;
    rep(i, 1, n)
        rep(j, 0, 9) rep(k, 0, d - 1)
            inc(f[i][(k + j) % d], f[i - 1][k]);

    int ret = 0;
    rep(i, 1, n - 1) rep(j, 1, 9) 
        inc(ret, f[i - 1][inv(j)]); 
    rep(i, 1, s[n] - 1) 
        inc(ret, f[n - 1][inv(i)]);
        
    int sum = s[n] % d;
    drep(i, n - 1, 1) {
        rep(j, 0, s[i] - 1)
            inc(ret, f[i - 1][inv(sum + j)]);
        sum = (sum + s[i]) % d;
    }
    if(!sum) ret ++;
    printf("%d
", ret);
}

int main() {
    scanf("%d", &d);
    scanf("%s", s + 1);
    n = strlen(s + 1); reverse(s + 1, s + n + 1);
    rep(i, 1, n) s[i] = s[i] - '0'; 
    Solve();
    return 0;
}