HDU 3555 Bomb

题目链接：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): 15270    Accepted Submission(s): 5515

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

 

Recommend

zhouzeyong

思路：经典的数位DP。自己不太懂这个知识点，参考了一位大神的博客，很详细，如下所示。

http://www.cnblogs.com/liuxueyang/archive/2013/04/14/3020032.html

附上AC代码：

#include <bits/stdc++.h>using namespace std;typedef long long ll;const int maxn = 25;ll dp[maxn][3];int num[maxn];ll n;void init(){dp[0][0] = 1;for (int i=1; i<maxn; ++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];}}int main(){init();int T;scanf("%d", &T);while (T--){scanf("%I64d", &n);ll ans = 0;bool ok = false;int len = 0;while (n){num[++len] = n%10;n /= 10;}for (int i=len; i>=1; --i){ans += dp[i-1][2]*num[i];if (ok)ans += dp[i-1][0]*num[i];if (!ok && num[i]>4)ans += dp[i-1][1];if (i+1<=len && num[i+1]==4 && num[i]==9)ok = true;}if (ok)++ans;printf("%I64d\n", ans);}return 0;}