Bomb
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 15072 Accepted Submission(s): 5441
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
题意：求1~n闭区间内含有“49”的数的个数
题解：
dp[i][2] 长度为i 含有“49”的个数
dp[i][1] 长度为i 不含有“49”但是高位为“9”的个数
dp[i][0] 长度为i 不含有“49”的个数
数组 a[i] 从低位到高位存储 n 的每一位数字。
dp[i][2]=dp[i-1][2]*10+dp[i-1][1]; //考虑第i位为“4” i-1位为“9”
dp[i][1]=dp[i-1][0];
dp[i][0]=dp[i-1][0]*10-dp[i-1][1];
对于n处理之前为什么要自增1
因为题目要求处理的是闭区间 可能自增1当作开区间处理
http://www.cnblogs.com/liuxueyang/archive/2013/04/14/3020032.html
1 /****************************** 2 code by drizzle 3 blog: www.cnblogs.com/hsd-/ 4 ^ ^ ^ ^ 5 O O 6 ******************************/ 7 //#include<bits/stdc++.h> 8 #include<iostream> 9 #include<cstring> 10 #include<cstdio> 11 #include<map> 12 #include<algorithm> 13 #include<queue> 14 #define ll __int64 15 using namespace std; 16 int t; 17 ll n; 18 ll a[65]; 19 ll dp[65][5]; 20 void init() 21 { 22 dp[0][0]=1; 23 for(int i=1; i<=22; i++) 24 { 25 dp[i][0]=10*dp[i-1][0]-dp[i-1][1]; 26 dp[i][1]=dp[i-1][0]; 27 dp[i][2]=10*dp[i-1][2]+dp[i-1][1]; 28 } 29 } 30 int main() 31 { 32 init(); 33 while(scanf("%d",&t)!=EOF) 34 { 35 for(int i=1; i<=t; i++) 36 { 37 scanf("%I64d",&n); 38 memset(a,0,sizeof(a)); 39 int len=1; 40 n++; 41 while(n) 42 { 43 a[len]=n%10; 44 n=n/10; 45 len++; 46 } 47 int flag=0; 48 int last=0; 49 ll ans=0; 50 for(int j=len; j>=1; j--) 51 { 52 ans+=dp[j-1][2]*a[j]; 53 if(flag) 54 ans+=dp[j-1][0]*a[j]; 55 if(!flag&&a[j]>4) 56 ans+=dp[j-1][1]; 57 if(last==4&&a[j]==9) 58 flag=1; 59 last=a[j]; 60 } 61 printf("%I64d ",ans); 62 } 63 } 64 return 0; 65 }