hdu 3555 Bomb

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
题目大概意思是：求1~N中含49的数，这题和2089是一个类型的题。但是注意的对于N+1处理。我求的是（0，n)的数，题目求的是【1，N】的数,然后去掉0.

#include<iostream>#include<algorithm>#include<cstring>#include<string>using namespace std;long long  dp[25][20];//此处要用 longlong int d[25];long long num(long long  nn){    int len=0;    long long ans=0;    while(nn)    {        d[++len]=nn%10;        nn/=10;     }      d[++len]=0;     for(int i=len;i>=1;i--)     {        for(int j=0;j<d[i];j++)        {            if(d[i+1]!=4||j!=9)            ans+=dp[i][j];         }         if(d[i]==9&&d[i+1]==4)            break;     }     return ans;}int main(){    int t;    cin>>t;    long long   n,x;    memset(dp,0,sizeof(dp));    dp[0][0]=1;    for(int i=1;i<=20;i++)//i位数     for(int j=0;j<10;j++)//最高位     for(int k=0;k<10;k++)//第2高位     if(j!=4||k!=9)      //j=4,k=9不同时成立     dp[i][j]+=dp[i-1][k];    while(t--)    {        cin>>n;        x=num(n+1);//不要49的数         cout<<n+1-x<<endl;    }    return 0; }