HDU 3555-digit DP

Bomb

Time limit:2000/1000 MS (java/others) Memory limit:131072/65536 K (java/others)
Total submission (s): 15072 Accepted Submission (s): 5441

Problem DescriptionThe 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 "a", 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?

Inputthe first line of input consists of an integer t (1 <= T <= 10000), indicating the number of test cases. For each test case, there'll be is an integer N (1 <= n <= 2^63-1) as the description.

The input terminates by end of file marker.

Outputfor each test case, output an integer indicating the final points of the power.

Sample Input3150500

Sample Output0115 HintFrom 1 to $, the numbers that include the Sub-sequence "is" 49 "," 149 "," 249 "," 349 "," 449 "," 490 "," 491 "," 492 "," 493 "," 49 4 "," 495 "," 496 "," 497 "," 498 "," 499 ", so the answer is 15.

Author[email protected]

Source2010 acm-icpc multi-university Training Contest (--host by WHU)

Test instructions: To find the number of "49" in the 1~n closed interval

Exercises

Dp[i][2] Length I contains the number of "49"

Dp[i][1] length I does not contain "49" but the number of "9" high

Dp[i][0] The length of I does not contain the number of "49"

Array A[i] stores every digit of n from low to high.

DP[I][2]=DP[I-1][2]*10+DP[I-1][1]; Consider the "4" i-1 bit "9" for the I-bit

DP[I][1]=DP[I-1][0];

DP[I][0]=DP[I-1][0]*10-DP[I-1][1];

Why do you want to increase your self by 1 before n processing

Because the problem is that the closed interval may be increased by 1 as an open interval processing

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

1 /******************************2 code by drizzle3 blog:www.cnblogs.com/hsd-/4 ^ ^    ^ ^5 o o6 ******************************/7 //#include <bits/stdc++.h>8#include <iostream>9#include <cstring>Ten#include <cstdio> One#include <map> A#include <algorithm> -#include <queue> - #definell __int64 the using namespacestd; - intT; - ll N; -ll a[ $]; +ll dp[ $][5]; - voidInit () + { Adp[0][0]=1; at      for(intI=1; i<= A; i++) -     { -dp[i][0]=Ten*dp[i-1][0]-dp[i-1][1]; -dp[i][1]=dp[i-1][0]; -dp[i][2]=Ten*dp[i-1][2]+dp[i-1][1]; -     } in } - intMain () to { + init (); -      while(SCANF ("%d", &t)! =EOF) the     { *          for(intI=1; i<=t; i++) $         {Panax Notoginsengscanf"%i64d",&n); -Memset (A,0,sizeof(a)); the             intlen=1; +n++; A              while(n) the             { +a[len]=n%Ten; -n=n/Ten; $len++; $             } -             intflag=0; -             intlast=0; thell ans=0; -              for(intJ=len; j>=1; j--)Wuyi             { theans+=dp[j-1][2]*A[j]; -                 if(flag) Wuans+=dp[j-1][0]*A[j]; -                 if(!flag&&a[j]>4) Aboutans+=dp[j-1][1]; $                 if(last==4&&a[j]==9) -flag=1; -last=A[j]; -             } Aprintf"%i64d\n", ans); +         } the     } -     return 0; $}

HDU 3555 Digit DP