HDU 3555 数位DP

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 11460    Accepted Submission(s): 4084

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

  
  

   
   3
1
50
500
  
  

 

Sample Output

  
  

   
   0
1
15

   
   
　　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。

  
  
  
  

   
   #include <iostream>
#include <cstdio>
#include<cstring>
#define LL long long
LL dp[21][3]; unsigned long long int n; int a[25];
int main(void){
//#ifndef ONLINE_JUDGE
  //freopen("hdu3555.in", "r", stdin);
//#endif
  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--){
    //cin >> n;
    scanf("%I64d", &n);
    int len = 0; memset(a, 0, sizeof(a));
    n++;
    while (n){
      a[++len] = n % 10; n /= 10;
    } LL ans = 0; int last = 0; bool flag = false;
    for (int i = len; i >= 1; --i){
      ans += (dp[i-1][2] * a[i]);
      if (flag) ans += dp[i-1][0] * a[i];
      if (!flag && a[i] > 4) {ans += dp[i-1][1];}
      if (last == 4 && a[i] == 9) {flag = true;}
      last = a[i];
    }
    //cout << ans << endl;
    printf("%I64d\n", ans);
  }


  return 0;
}