Bomb(HDU 3555 数位DP)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 20230    Accepted Submission(s): 7565

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

3150500

 

Sample Output

0115
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.
 

 
//题意：给一个数字n，问1-n里有几个数包含49.

//思路：数位DP：

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。

状态转移方程：

dp[i][0] = dp[i-1][0] * bit[i] - dp[i-1][1];

dp[i][1]=dp[i-1][1];

dp[i][2]=dp[i-1][2]*bit[i]+dp[i-1][1];

#include <iostream>#include <cstdio>#include <cmath>#include <cstring>#include <algorithm>using namespace std;typedef long long ll;ll n;int bit[25];ll dp[25][5]; int main(){    int T;    scanf("%d",&T);    memset(dp,0,sizeof(dp));    dp[0][0]=1;    for(int i=1;i<21;i++)    {        dp[i][0]=dp[i-1][0]*10-dp[i-1][1];        dp[i][1]=dp[i-1][0];        dp[i][2]=dp[i-1][2]*10+dp[i-1][1];    }    while(T--)    {        scanf("%I64d",&n);        memset(bit,0,sizeof(bit));        ll x=n+1;        int len=0;        while(x)        {            bit[++len]=x%10;            x=x/10;        }        ll ans=0;        bool flag=false;        int last=0;        for(int i=len;i>=1;i--)        {            ans+=(dp[i-1][2]*bit[i]);            if(flag)                ans+=dp[i-1][0]*bit[i];            if(!flag&&bit[i]>4)                ans+=dp[i-1][1];            if(last==4&&bit[i]==9)                flag=true;            last=bit[i];        }        printf("%I64d\n",ans);    }    return 0;}