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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.
思路:http://blog.csdn.net/winkloud/article/details/7866520
#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\n",solve(r+1));   }   return 0;}

 

HDU 3555 Bomb