codeforces 628D 数位DP

codeforces 628D

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题意：

$给定区间[a,b]，限制条件：$

• $能被m整除$
• $偶数位全部为d且奇数位不能为d$

$问区间内满足以上两个条件的数的个数，结果对1e9+7取模。$

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题解：

$dp[pos][num]表示查找到第pos位，对m取模余数为num的合法数的个数。$

• $记忆化搜索，dfs(pos-1,(num\ast 10+i)\%m,limit\&\&i==bit[pos])$

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#include 
using namespace std;
typedef long long ll;
const int mod = 1e9+7;
const int N = 2001;
int m, d, len, bit[N];
ll dp[N][N];

ll dfs(int pos, int num, bool limit){
    if(pos == -1){
        return num == 0;
    }
    if(!limit && dp[pos][num] != -1){
        return dp[pos][num];
    }
    int up = (limit ? bit[pos] : 9);
    ll res = 0;
    for(int i = 0 ; i <= up ; i++){
        if(((len-pos)%2 == 0 && i != d) || ((len-pos)%2 == 1 && i == d)) continue;
        res += dfs(pos-1, (num*10+i)%m,  limit && i == bit[pos])%mod;
    }
    if(!limit){
        dp[pos][num] = res;
    }
    return res;
}

ll count(string s){
    len = s.size();
    for(int i = 0 ; i < len ; i++){
        bit[i] = s[len-i-1]-'0';
    }
    return dfs(len-1, 0, true);
}

int main() {
    memset(dp, -1, sizeof(dp));
    cin >> m >> d;
    string sl, sr;
    cin >> sl >> sr;
    ll flag = 1, num = 0;
    for(int i = 0 ; i < sl.size() && flag ; i++){
        num = (num*10+sl[i]-'0')%m;
        if(((i+1)%2 == 0 && sl[i]-'0' != d) || ((i+1)%2 == 1 && sl[i]-'0' == d)){
            flag = 0;
        }
    }
    if(flag){
        flag = (num ? 0 : 1);
    }
    cout << ((count(sr)-count(sl)+flag)%mod+mod)%mod << endl;
    return 0; 
}