Educational Codeforces Round 8 D - Magic Numbers 数位DP

题意很简单，偶数位为d，奇数位不为d，且能被m整除的数，叫d-magic，给定d，m，a，b，求在[a,b]上d-magic的个数。

注意 1 既是 0-magic 又是 2/3/4/5/6/7/8/9-magic

还有值得注意的是，给定的a，b 数位相同。 一开始没用到这个条件，写得挺麻烦～

#include<stdio.h>
#include<string.h>
#include<ctype.h>
#include<math.h>
#include<string>
#include<set>
#include<map>
#include<vector>
#include<queue>
#include<algorithm>
#include<ctime>
using namespace std;
void fre(){freopen("t.txt","r",stdin);}
template <class T1, class T2>inline void gmax(T1 &a, T2 b) { if (b>a)a = b; }
template <class T1, class T2>inline void gmin(T1 &a, T2 b) { if (b<a)a = b; }
#define ls o<<1
#define rs o<<1|1
#define MS(x,y) memset(x,y,sizeof(x))
#define debug(x) printf("%d",x);
typedef long long LL;
typedef unsigned long long UL;
typedef unsigned int UI;
const int INF = 1<<30;
const int MAXN = 105;
const double eps = 1e-8;
const double pi = acos(-1.0);
const LL M = 1e9+7;

LL dp[2005][2005][2];
int num[2005];
char s[2005];
LL m,d,all;
LL dfs(int top, int s, bool e, int flag)//top为第几位数，s为模m的余数，e为是否到上限，flag表示偶数位还是奇数位
{
if (top==0) return (s==0 ) ;
if (!e && dp[top][s][flag]!=-1) return dp[top][s][flag];
int i;
LL res = 0;
int u = e?num[top]:9;
if(flag) {if(u >= d) res = dfs(top-1,(s*10+d)%m, e&&d==u,flag^1);}
else
{
for (i = 0; i <= u; ++i)
{
if(i == d) continue;//由于a，b数位相同，所以不用考虑前导0的情况。这里为了代码简单就没改，反正都会减掉的。但是当d = 0 的时候，算出来的并不是
准确的d-magic 个数
res = (res + dfs(top-1, (s*10+i)%m, e&&i==u, flag^1))%M;
}
}
return e?res:dp[top][s][flag]=res;
}
bool check()
{
int flag = 1,mo = 0;
for(int i = all; i >= 1; --i)
{
if(!flag && num[i]!=d) return 0;
else if(flag && num[i] == d) return 0;
mo = ((mo*10)+num[i])%m;
flag^=1;
}
if(mo) return 0;
else return 1;
}
int main()
{
// fre();
int i;
MS(dp,-1);
scanf("%I64d%I64d",&m,&d);

scanf("%s",s); all = strlen(s);
for(i = 0; s[i]; ++i) num[all-i] = s[i]-'0';
LL tem1 = dfs(all,0,1,0);
if(check()) tem1-=1;

scanf("%s",s);
for(i = 0; s[i]; ++i) num[all-i] = s[i]-'0';
LL tem2 = dfs(all,0,1,0);

printf("%I64d\n",((tem2 - tem1)%M+M)%M );
}