Bomb

**Time Limit: 2000/1000 MS (Java/others) memory limit: 131072/65536 K (Java/Others)**

Total submission (s): 7921 accepted submission (s): 2778

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 between des the sub-sequence "49", the power of the blast wocould 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 will be 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

*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 [email protected]

Source 2010 ACM-ICPC multi-university training Contest (12) -- host by whu question: Give You A number N, You need to count 1 to n appears contains 49 numbers: for example, 498,549, 49 ..... for this question: when I see a blog referencing this image and think it is clear, I will reference it .. our numerical analysis of the length of the I-1 is undoubtedly so concentrated (of course only around 49 Wow) first analysis: the length of the I-1 so for the I Length header is 49, the format must be: 49 ****? 49 ****(? It may be 9) the first letter of guarantee 9, so its format must be: 9 *****? 9 *****(? It may be 4) header position 49, so its format is :*******? *******(? May be 9) We may use DP [I] [2] to indicate that the header is 49, and DP [I] [1] to indicate that the header is 9, DP [I] [0] indicates that the header is not 49, so we can find a rule: DP [I-1] [2] Move one digit forward, that is, the original bit becomes ten, the shift from 10 to bits. The formation of DP [I] [2], but note: When DP [I-1] [2], actually I said above ,? It may be 9, so when we move one forward ,? The possibility of being 9 is removed. So DP [I-1] [2] * 10 (when moving one digit) needs to subtract the pattern DP [I-1] [1] starting with 9, so get: (1) DP [I] [2] = DP [I-1] [2] * 10-dp [I-1] [1]; for the I-bit header is 9 then the back only needs to meet not 49 can, exactly meet DP [I] [0]; (2) So DP [I] [1] = d [I-1] [0]; you can also analyze the content of the header that is not 49... DP [I] [0] = DP [I-1] [0] * 10 + dp [I-1] [1]; so a preprocessing equation is obtained: DP [I] [2] = DP [I-1] [2] * 10-dp [I-1] [1]; DP [I] [1] = d [I-1] [0]; DP [I] [0] = DP [I-1] [0] * 10 + dp [I-1] [1]; Code: For details, see the code:

1 // # define local 2 # include <cstdio> 3 # include <cstring> 4 # define ll _ int64 5 using namespace STD; 6 const int maxn = 25; 7 ll dp [maxn] [3] = {0}; 8 int NN [maxn]; 9 int main () 10 {11 12 # ifdef local13 freopen ("test. in "," r ", stdin); 14 # endif15 int cas, I; 16 LL N; 17 scanf (" % d ", & CAS ); 18/* Common Mode preprocessing of digital DP */19 DP [0] [0] = 1; 20 for (I = 1; I <= 20; I ++) 21 {22 DP [I] [0] = DP [I-1] [0] * 10-dp [I-1] [1]; 23 DP [I] [1] = DP [I-1] [0]; 24 DP [I] [2] = DP [I-1] [2] * 10 + dp [I-1] [1]; 25} 26 while (CAS --) 27 {28 scanf ("% i64d", & N); 29 I = 0; 30 N + = 1; 31 memset (NN, 0, sizeof (NN )); 32 While (n> 0) 33 {34 NN [++ I] = n % 10; 35 N/= 10; 36} 37 ll ans = 0; 38 bool tag = 0; 39 int num = 0; 40 for (; I> = 1; I --) 41 {42 ans + = DP [I-1] [2] * NN [I];/* calculate the number starting with 49 */43 If (TAG) {44 ans + = DP [I-1] [0] * NN [I];/* when the current surface appears 49, then any number that appears later must be calculated */45} 46 If (! Tag & NN [I]> 4) 47 {48 ans + = DP [I-1] [1];/* If the start of 49 does not appear, as long as the header is greater than 5, then there must be a 49 */49} 50 if (num = 4 & NN [I] = 9) 51 tag = 1; 52 num = nn [I]; 53} 54 printf ("% i64d \ n", ANS); 55} 56 return 0; 57}

View code

HDU --- (3555) bomb (Digital dp (Getting Started ))