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Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 10967 Accepted Submission(s): 3890
1 #include <bits/stdc++.h> 2 using namespace std; 3 typedef long long LL; 4 const int maxn = 21; 5 LL dp[maxn][3]; 6 int b[maxn]; 7 LL dfs(int p,int st,bool flag){ 8 if(!p) return st == 2; 9 if(flag && dp[p][st] != -1) return dp[p][st]; 10 int u = flag?9:b[p]; 11 LL ret = 0; 12 for(int i = 0; i <= u; ++i){ 13 if(st == 1 && i == 9 || st == 2) ret += dfs(p-1,2,flag||(i < u)); 14 else if(i == 4) ret += dfs(p-1,1,flag||(i < u)); 15 else ret += dfs(p-1,0,flag||(i < u)); 16 } 17 if(flag) dp[p][st] = ret; 18 return ret; 19 } 20 LL solve(LL x){ 21 int len = 0; 22 while(x){ 23 b[++len] = x%10; 24 x /= 10; 25 } 26 return dfs(len,0,false); 27 } 28 int main(){ 29 memset(dp,-1,sizeof dp); 30 int kase; 31 scanf("%d",&kase); 32 while(kase--){ 33 LL tmp; 34 scanf("%I64d",&tmp); 35 printf("%I64d\n",solve(tmp)); 36 } 37 return 0; 38 }
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原文地址:http://www.cnblogs.com/crackpotisback/p/4782213.html