Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 11804 Accepted Submission(s): 4212
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?
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.
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
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
【思路】
数位DP。
预处理：
f[i][0] 表示i位数中不含49的数的数目
f[i][1] 表示i位数中不含49且最高位为9的数的数目
f[i][2] 表示i位数中含49的数的数目
根据n的每一位统计ans，具体见代码。
【代码】
1 #include<cstdio> 2 using namespace std; 3 4 typedef long long LL; 5 LL f[25][3]; 6 /* 7 f[i][0] i位数 无49存在 8 f[i][1] i位数 无49存在且末尾为9 9 f[i][2] i位数 有49 10 */ 11 int b[25]; 12 13 void init() { 14 f[0][0]=1; f[0][1]=f[0][2]=0; 15 for(int i=1;i<25;i++) { 16 f[i][0]=10*f[i-1][0]-f[i-1][1]; 17 f[i][1]=f[i-1][0]; 18 f[i][2]=10*f[i-1][2]+f[i-1][1]; 19 } 20 } 21 22 int main() { 23 int T; LL n; 24 scanf("%d",&T); 25 init(); 26 while(T--) { 27 scanf("%I64d",&n); 28 int len=0; 29 while(n) 30 b[++len]=n%10 , n/=10; 31 b[len+1]=0; 32 LL ans=0; bool flag=0; 33 for(int i=len;i;i--) { 34 ans += b[i]*f[i-1][2]; // + ...49... 35 if(flag) ans += f[i-1][0]*b[i]; // + 49...(无49...) 36 else if(b[i]>4) ans += f[i-1][1]; // + 4(9 无49) 37 if(b[i+1]==4 && b[i]==9) flag=1; 38 } 39 if(flag) ans++; //算上自身 40 printf("%I64d ",ans); 41 } 42 return 0; 43 }