Bomb（数位Dp）

Bomb

Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 10373 Accepted Submission(s): 3657

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

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.

Author

fatboy_cw@WHU

#include<cstdio>
#include<iostream>
#include<cstring>
#define ll long long
using namespace std;
ll dp[21][3];
void init()
{
    int i;
    memset(dp,0,sizeof(dp));
    dp[0][0]=1;
    /*
    dp[i][0]代表长度为 i 并且不含有49的数字的个数；
　　dp[i][1]代表长度为 i 并且不含有49，但是最高位是9的数字的个数；
　　dp[i][2]代表长度为 i 并且含有49的数字的个数。
    */
    for(i=1;i<21;i++)
    {
        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];// 表示长度为 i 的不含有49的数字的个数等于
       // 长度为 i - 1 的不含有49的数字的个数*当前的数字
        dp[i][1]=dp[i-1][0];//表示长度为 i 的并且不含有49同时最高位是9的数字的个数等于，
       //长度为 i - 1 的不含有49的数字的个数，
       //因为只要在它的高一位加上一个9就可以了。
        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];
         /*
          表示长度为 i 的含有49的数字的个数等于，
          长度为 i - 1 的数字的个数*当前的数字
          ，再加上长度为 i - 1 的
          并且不含有49同时最高位是9的数字的个数，因为这个时候，
          只要在高一位加上一个4就可以了，
          这样在最高的两位就组成了一个49。*/
    }
}
int main()
{
    int i,t,len,last,flag,a[22];
    ll ans,n;
    init();
    cin>>t;
    while(t--)
    {
        memset(a,0,sizeof(a));
        last=flag=ans=0;//flag用于判断这个数之前的数是否是49
        cin>>n;
        n++;
        len=1;
        while(n)
        {
            a[len++]=n%10;
            n/=10;
        }
        len--;
        for(i=len;i>=1;i--)
        {
            ans+=dp[i-1][2]*a[i];//在i之前的一位所有含有49数的种类*当前数
            if(flag)//如果这个数与前一个数组合成49
            {
                ans+=dp[i-1][0]*a[i];//这个数的种类就是下一位所有的非49数*当前数
                //比如4960，应该是6*1=6;
            }
            if(!flag&&a[i]>4)
            {
                ans+=dp[i-1][1];
            }
            if(last==4&&a[i]==9)
            {
                flag=true;
            }
            last=a[i];
        }
        printf("%I64d\n",ans);
    }
}