HDU 3555 Bomb [数位] - c++编程基础

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HDU 3555 Bomb [数位]

2015-11-21 01:04:48 来源: 作者: 【大中小】浏览:1次

Description

The counter-terrorists found a time bomb in the dust. But this time the terrorists improve on the time bomb. The number sequence of the time bomb counts from 1 to N. If the current number sequence includes the sub-sequence "49", the power of the blast would add one point.
Now the counter-terrorist knows the number N. They want to know the final points of the power. Can you help them?

Input

The first line of input consists of an integer T (1 <= T <= 10000), indicating the number of test cases. For each test case, there will be an integer N (1 <= N <= 2^63-1) as the description.

The input terminates by end of file marker.

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

Sample Output

 0
1
15

Hint

From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.

Analyse:

同 “不要62”，数位模板！！

但是脑残，，，，dp没开long long；

Code：

#include
  
   
#include
   
     #include
    
      #include
     
       #include
      
        #include
       
         #include
        
          #include
         
           #include
          
            #include
           
             #include
             #include
             
               #define INF 0x7fffffff #define SUP 0x80000000 #define mem(a,b) memset(a,b,sizeof(a)) using namespace std; typedef __int64 LL; const int N=100007; LL dp[30][10][2]; int digit[30]; LL dfs(int pos,int pre,int en,int limit) { if(pos==-1) return en; if(!limit&&dp[pos][pre][en]!=-1) return dp[pos][pre][en]; int last=limit?digit[pos]:9; LL ret=0; for(int i=0;i<=last;i++) { ret+=dfs(pos-1,i,en||(pre==4&&i==9),limit&&i==last); } if(!limit) dp[pos][pre][en]=ret; return ret; } LL solve(LL x) { int cnt=0; while(x) { digit[cnt++]=x%10; x/=10; } return dfs(cnt-1,0,0,1); } int main() { int T;scanf("%d",&T); LL n; mem(dp,-1); while(T--) { scanf("%I64d",&n); printf("%I64d\n",solve(n)); } return 0; }


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