CodeForces 628 D Magic Numbers 数位DP

Magic Numbers

题意：

题意比较难读：首先对于一个串来说， 如果他是d-串， 那么他的第偶数个字符都是是d，第奇数个字符都不是d。

然后求[L, R]里面的多少个数是d-串，且是m的倍数。

题解：

数位dp。

dp[x][y]代表的是余数为x， 然后剩下的长度是y的情况的方案数是多少。

代码：

#include<bits/stdc++.h>
using namespace std;
#define Fopen freopen("_in.txt","r",stdin); freopen("_out.txt","w",stdout);
#define LL long long
#define ULL unsigned LL
#define fi first
#define se second
#define pb push_back
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define lch(x) tr[x].son[0]
#define rch(x) tr[x].son[1]
#define max3(a,b,c) max(a,max(b,c))
#define min3(a,b,c) min(a,min(b,c))
typedef pair<int,int> pll;
const int inf = 0x3f3f3f3f;
const int _inf = 0xc0c0c0c0;
const LL INF = 0x3f3f3f3f3f3f3f3f;
const LL _INF = 0xc0c0c0c0c0c0c0c0;
const LL mod =  (int)1e9+;
const int N = 2e3 + ;
int dp[N][N];
/// left length
char s[N];
int m, d;
///      length, yu, lim
int dfs(int len, int yu, int lim, int g){
    if(!lim && dp[yu][len] != -) return dp[yu][len];
    if(len == ){
        return !yu;
    }
    int up = ;
    if(lim) up = s[len] - '';
    LL tmp = ;
    for(int i = ; i <= up; ++i){
        if(g){
            if(i != d) continue;
        }
        else {
            if(i == d) continue;
        }
        tmp += dfs(len-, (yu*+i)%m, lim && (i==up), g^);
    }
    tmp %= mod;
    if(!lim) dp[yu][len] = tmp;
    return tmp;
}
bool check(int len){
    int yu = , g = ;
    for(int i = len; i >= ; --i){
        if(g && s[i] != '' + d) return false;
        if(!g && s[i] == '' + d) return false;
        yu = (yu *  + s[i] - '') % m;
        g ^= ;
    }
    return yu == ;
}
int main(){
    memset(dp, -, sizeof(dp));
    scanf("%d%d", &m, &d);
    scanf("%s", s+);
    int len1 = strlen(s+);
    reverse(s+, s++len1);
    LL ans = check(len1) - dfs(len1, , , );
    scanf("%s", s+);
    len1 = strlen(s+);
    reverse(s+, s++len1);
    ans += mod + dfs(len1, , , );
    ans %= mod;
    cout << ans << endl;
    return ;
}