思路：数位DP
1 #include<cstdio> 2 #include<cstring> 3 int bit[10],dp[10][10][13][2]; 4 int dfs(int pos,int pre,int mod,bool h,bool inf) 5 { 6 if(pos==-1) return h&&!mod; 7 if(!inf&&dp[pos][pre][mod][h]!=-1) return dp[pos][pre][mod][h]; 8 int i,ans=0; 9 int e=inf?bit[pos]:9; 10 for(i=0;i<=e;i++){ 11 if(pre==1&&i==3) ans+=dfs(pos-1,i,(mod*10+i)%13,1,inf&&(i==bit[pos])); 12 else ans+=dfs(pos-1,i,(mod*10+i)%13,h,inf&&(i==bit[pos])); 13 } 14 if(!inf) dp[pos][pre][mod][h]=ans; 15 return ans; 16 } 17 int cal(int n) 18 { 19 int pos=0; 20 while(n){ 21 bit[pos++]=n%10; 22 n/=10; 23 } 24 return dfs(pos-1,0,0,0,1); 25 } 26 int main() 27 { 28 int n,ans,m,k; 29 memset(dp,-1,sizeof(dp)); 30 while(scanf("%d",&n)!=EOF){ 31 printf("%d ",cal(n)); 32 } 33 }