codeforces 628D. Magic Numbers 数位dp

题目链接

给两个数m, d. 两个数a, b。 a, b长度小于2000， 长度相等。求在a, b之间的数x， x%m==0， 并且从高位往低位数， 奇数位的数全部不等于d， 偶数为的数全都等于d， 求这样的数的个数。

数位dp， 通常的都是从后往前dfs， 但是这个题因为有第二个条件， 所以应该从前往后dfs， 具体的dfs过程看代码。

#include <iostream>
#include <vector>
#include <cstdio>
#include <cstring>
#include <algorithm>
#include <cmath>
#include <map>
#include <set>
#include <string>
#include <queue>
#include <stack>
#include <bitset>
using namespace std;
#define pb(x) push_back(x)
#define ll long long
#define mk(x, y) make_pair(x, y)
#define lson l, m, rt<<1
#define mem(a) memset(a, 0, sizeof(a))
#define rson m+1, r, rt<<1|1
#define mem1(a) memset(a, -1, sizeof(a))
#define mem2(a) memset(a, 0x3f, sizeof(a))
#define rep(i, n, a) for(int i = a; i<n; i++)
#define fi first
#define se second
typedef pair<int, int> pll;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int mod = 1e9+7;
const int inf = 1061109567;
const int dir[][2] = { {-1, 0}, {1, 0}, {0, -1}, {0, 1} };
int m, d, len;
char s[2005];
ll dp[2005][2005];
ll dfs(int pos, int md, int fp) {
if(pos == len+1) {
return md == 0;
}
if(~dp[pos][md]&&!fp) {
return dp[pos][md];
}
int ed = fp?s[pos]-'0':9;
ll ret = 0;
for(int i = 0; i<=ed; i++) {
if(pos%2&&i==d)
continue;
if(pos%2==0&&i!=d)
continue;
ret = (ret + dfs(pos+1, (md*10+i)%m, fp&&i == ed))%mod;
}
return fp?ret:dp[pos][md] = ret;
}
int main()
{
mem1(dp);
cin>>m>>d;
scanf("%s", s+1);
len = strlen(s+1);
int tmp = 0, flag = 0;
for(int i = 1; i<=len; i++) {
tmp = tmp*10+s[i]-'0';
tmp %= m;
if(i%2&&s[i]-'0' == d)
flag = 1;
if(i%2==0&&s[i]-'0'!=d)
flag = 1;
}
if(tmp!=0)
flag = 1;
ll ans1 = dfs(1, 0, 1);
if(!flag)
ans1--;
scanf("%s", s+1);
ll ans2 = dfs(1, 0, 1);
cout<<(ans2-ans1+mod)%mod;
return 0;
}