Educational Codeforces Round 8 D. Magic Numbers

Magic Numbers

题意:给定长度不超过2000的a,b;问有多少个x(a<=x<=b)使得x的偶数位为d,奇数位不为d;且要是m的倍数，结果mod 1e9+7;

直接数位DP;前两维的大小就是mod m的大小，注意在判断是否f[pos][mod] != -1之前，要判断是否为边界，否则会出现重复计算；

#include<bits/stdc++.h>
using namespace std;
#define inf 0x3f3f3f3f
#define MS1(a) memset(a,-1,sizeof(a))
const int N = ,MOD = 1e9+;
char a[N],b[N];
int f[N][N][],n,m,d;
int dfs(char* s,int p,int mod,int on = )
{
    if(p == n) return mod == ;
    int& ans = f[p][mod][on];
    if(ans != -) return ans;
    ans = ;
    for(int i = ;i <= (on?s[p] - '':);i++){
        if(p% && i != d) continue;
        if(p% ==  && i == d) continue;
        (ans += dfs(s,p+,(mod*+i)%m,on && i == s[p] - '')) %= MOD;
    }
    return ans;
}
int calc(char* s)
{
    MS1(f);
    return dfs(s,,);
}
int main()
{
    scanf("%d%d",&m,&d);
    scanf("%s%s",a,b);
    n = strlen(a);
    for(int i = n-;i >= ;i--){ // a - 1;
        if(a[i] == '') a[i] = '';
        else{a[i]--;break;}
    }
    printf("%d",(calc(b)-calc(a)+MOD)%MOD);
}