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  • HDU 3555 Bomb

    http://acm.hdu.edu.cn/showproblem.php?pid=3555

    Bomb

    Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
    Total Submission(s): 7763    Accepted Submission(s): 2717


    Problem 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
    3 1 50 500
     

     

    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.
    #include<iostream>
    #include<cstring>
    #include<cstdio>
    using namespace std;
    __int64 dp[25][3];
    int bit[25];
    void init()
    {
        memset(dp,0,sizeof(dp));
        int i;
        dp[0][0]=1;
        for(i=1;i<=20;i++)
          {
              dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
              //dp[i][0] 表示i位数字中不含49的数字的个数  
              dp[i][1]=dp[i-1][0];
              //dp[i][1] 表示i位数字中以9开头的数字的个数  
              dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
              ////dp[i][2] 表示i位数字中含有49的数字的个数 
        }
    }
    __int64 solve(__int64 n)
    {
      memset(bit,0,sizeof(bit));
        int len=1,i;
        __int64 temp=n;
        while(temp)
        {
            int r=temp%10;
            bit[len++]=r;
            temp=temp/10;
        }
        int flag=0;
        __int64 ans=0;
        bit[len]=0;
        for(i=len-1;i>=1;i--)
        {
             ans+=dp[i-1][2]*bit[i];
             if(flag)
              {
                  ans+=dp[i-1][0]*bit[i];
              }
            //if(!flag&&bit[i+1]==4&&bit[i]>=9)
            //{
            //    ans+=dp[i][1];
    
           // }
            if(!flag&&bit[i]>4)
                 {
                    ans+=dp[i-1][1];
                 }
            if(bit[i+1]==4&&bit[i]==9)
                 flag=1;
          }
         return ans;
    
    }
    int main()
    {
        int T;
        __int64 r;
        scanf("%d",&T);
       init();
       while(T--)
       {
           scanf("%I64d",&r);
    
           printf("%I64d
    ",solve(r+1));
       }
       return 0;
    }
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  • 原文地址:https://www.cnblogs.com/cancangood/p/3946577.html
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