codeforces 628D Magic Numbers
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题目链接:【codeforces 628D】
d-magic:从左往右数(从1开始标记),数字num的第偶数个数字是d,其他不是d。比如17273是7-magic,8的0-magic,34是4-magic
输入m,d,a,b(1<=m<=2000, 0<=d<=9, a,d的位数相同,并且位数小于2000),找x的个数(结果对1e9+7取余),x必须满足以下条件:
1、a<=x<=b && x%m==0
2、x是d-magic
#include <bits/stdc++.h>using namespace std;#define ll __int64const int inf=2010;const int mod=1e9+7;ll dp[inf][inf];int m, d, bit[inf], len;ll dfs(int pos, int res, int lim){if(pos==len+1) return res==0;if(!lim && dp[pos][res]!=-1) return dp[pos][res];int end=lim?bit[pos]:9;ll ans=0;for(int i=0; i<=end; i++){if(pos%2==1 && i==d) continue;if(pos%2==0 && i!=d) continue;ans = (ans+dfs(pos+1, (res*10+i)%m, lim&&i==end))%mod; }if(!lim) dp[pos][res]=ans;return ans;}ll solve(string a){for(int i=0; i<len; i++){bit[i+1]=a[i]-'0';}return dfs(1, 0, 1);}ll check(string a){int res=0;for(int i=1; i<=len; i++){if(i%2==0 && a[i-1]-'0'!=d) return 0;if(i%2==1 && a[i-1]-'0'==d) return 0;res=res*10+(a[i-1]-'0');res%=m;}if(res) return 0;return 1;}int main(){memset(dp, -1, sizeof(dp));scanf("%d%d", &m, &d);string a, b;cin>>a>>b;len=a.size();printf("%I64d\n", (solve(b)-solve(a)+check(a)+mod)%mod);return 0;}
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