HDU3555：Bomb（数位dp入门）

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 16873    Accepted Submission(s): 6182

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.
 

 

Author

fatboy_cw@WHU

 

Source

2010 ACM-ICPC Multi-University Training Contest（12）——Host by WHU

题意：计算1~N有几个数包含“49”。

思路：计算不包含的再减一下就行， long long 输入输出用I64d

# include <stdio.h># include <string.h>long long a[25], dp[25][2]={0};long long  dfs(int pos, int sta, bool limit){    if(pos == -1) return 1;    if(!limit && dp[pos][sta] != -1) return dp[pos][sta];    int up=limit?a[pos]:9;    long long ans = 0;    for(int i=0; i<=up; ++i)    {        if(sta && i==9)            continue;        ans += dfs(pos-1, i==4, limit&&i==a[pos]);    }    if(!limit) dp[pos][sta] = ans;    return ans;}long long solve(long long num){    int pos = 0;    while(num)    {        a[pos++] = num%10;        num /= 10;    }    return dfs(pos-1, 0, true);}int main(){    memset(dp, -1, sizeof(dp));    int  t;    long long n;    scanf("%d",&t);    while(t--)    {        scanf("%I64d",&n);        printf("%I64d\n",n-solve(n)+1);    }    return 0;}