Educational Codeforces Round 8 D. Magic Numbers （数位DP）-优快云博客

本文链接：https://blog.youkuaiyun.com/Ezereal/article/details/52385501

探讨了如何使用数位动态规划（DP）方法解决在指定区间内寻找满足特定条件的魔法数（偶数位为d，奇数位不为d且能被m整除）的数量问题。

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题意：给一个m一个d，一个字符串a和b，问在[a,b]范围内，有多少个可以整除m的魔法数，魔法数的定义是，偶数位上都是d，奇数位上都不是d。

典型的数位DP

#include <set>
#include <map>
#include <stack>
#include <queue>
#include <deque>
#include <cmath>
#include <vector>
#include <string>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define L(i) i<<1
#define R(i) i<<1|1
#define INF  0x3f3f3f3f
#define pi acos(-1.0)
#define eps 1e-9
#define maxn 100100
#define MOD 1000000007

long long dp[2020][2020];
int bit[2020];
int m,d;
int len;

long long dfs(int pos,int st,int zero,int flag,int odd)
{
    //printf("%d %d %d %d %d\n",pos,st,zero,flag,odd);
    if(pos == len)
        return st == 0;
    if(flag && dp[pos][st] != -1)
        return dp[pos][st];
    int u = flag?9:bit[pos];
    long long ans = 0;
    if(!odd)
    {
        for(int i = 0; i <= u; i++)
            if(i != d)
                ans += dfs(pos+1,(st*10+i)%m,zero||i,flag||i<u,odd^1);
    }
    else
    {
        if(flag)
            ans += dfs(pos+1,(st*10+d)%m,zero||d,flag||d<u,odd^1);
        else if(d <= u)
            ans += dfs(pos+1,(st*10+d)%m,zero||d,flag||d<u,odd^1);
    }
    ans %= MOD;
    if(flag)
        dp[pos][st] = ans;
    return ans;
}

long long solve(char s[])
{
    len = strlen(s);
    for(int i = 0; i < len; i++)
        bit[i] = s[i] - '0';
    return dfs(0,0,0,0,0);
}

int main()
{
    //freopen("in.txt","r",stdin);
    //freopen("out.txt","w",stdout);
    int t,C = 1;
    //scanf("%d",&t);
    while(scanf("%d%d",&m,&d) != EOF)
    {
        memset(dp,-1,sizeof(dp));
        char s1[2020],s2[2020];
        scanf("%s%s",s1,s2);
        int len = strlen(s1);
        int flag = 1,u = 0,cnt = 0;
        for(int i = 0; i < len; i++)
        {
            u ^= 1;
            if(u & 1 && s1[i] == '0'+d)
                flag = 0;
            if(!(u&1) && s1[i] != '0'+d)
                flag = 0;
            cnt = cnt * 10 + s1[i] - '0';
            cnt %= m;
        }
        if(cnt)
            flag = 0;
        //printf("%lld %lld\n",solve(s1),solve(s2));
        printf("%lld\n",(solve(s2)-solve(s1)+flag+MOD)%MOD);
    }
    return 0;
}