CodeForces 628D Magic Numbers (数位DP)

Posted 2020-08-14 dwtfukgv

题意：求给定区间内偶数位置全是d并且是m的倍数的数的个数。

析：一开始以为是偶数有的是d，有的不是，然后还没有看到区间的长度相等，这个是十分重要的，要不然开不出数组。

dp[i][j] 表示前 i 位，取模m为 j。

代码如下：

#pragma comment(linker, "/STACK:1024000000,1024000000")
#include <cstdio>
#include <string>
#include <cstdlib>
#include <cmath>
#include <iostream>
#include <cstring>
#include <set>
#include <queue>
#include <algorithm>
#include <vector>
#include <map>
#include <cctype>
#include <cmath>
#include <sstream>
#include <stack>
//#include <tr1/unordered_map>
#define freopenr freopen("in.txt", "r", stdin)
#define freopenw freopen("out.txt", "w", stdout)
using namespace std;
//using namespace std :: tr1;

typedef long long LL;
typedef pair<int, int> P;
const int INF = 0x3f3f3f3f;
const double inf = 0x3f3f3f3f3f3f;
const LL LNF = 0x3f3f3f3f3f3f;
const double PI = acos(-1.0);
const double eps = 1e-8;
const int maxn = 2e3 + 5;
const int mod = 1e9 + 7;
const int N = 1e6 + 5;
const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1};
const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1};
const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"};
inline LL gcd(LL a, LL b){  return b == 0 ? a : gcd(b, a%b); }
int n, m;
const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31};
inline int Min(int a, int b){ return a < b ? a : b; }
inline int Max(int a, int b){ return a > b ? a : b; }
inline LL Min(LL a, LL b){ return a < b ? a : b; }
inline LL Max(LL a, LL b){ return a > b ? a : b; }
inline bool is_in(int r, int c){
    return r >= 0 && r < n && c >= 0 && c < m;
}
LL dp[maxn][maxn];
int a[maxn];
char s[maxn], t[maxn];
int k, len;

LL dfs(int pos, int val, bool ok){
    if(pos == len)  return val == 0;
    LL &ans = dp[pos][val];
    if(!ok && ans >= 0)  return ans;

    LL res = 0;
    int n = ok ? a[pos] : 9;
    for(int i = 0; i <= n; ++i){
        if((pos&1) && i != m)  continue;
        if(!(pos&1) && i == m)  continue;
        res = (res + dfs(pos+1, (val*10+i)%k, ok && i == n)) % mod;
    }
    return ok ? res : ans = res;
}

LL solve(char *s){
    for(len = 0; s[len]; ++len)
        a[len] = s[len] - ‘0‘;
    return dfs(0, 0, true);
}

bool judge(char *s){
    int val = 0;
    for(int i = 0; s[i]; ++i){
        if((i&1) && s[i] - ‘0‘ != m)  return false;
        else if(!(i&1) && s[i] - ‘0‘ == m)  return false;
        val = (val * 10 + s[i] - ‘0‘) % k;
    }
    return val == 0;
}

int main(){
    while(scanf("%d %d", &k, &m) == 2){
        memset(dp, -1, sizeof dp);
        scanf("%s %s", s, t);
        cout << (solve(t) - solve(s) + judge(s) + mod) % mod << endl;
    }
    return 0;
}