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
#include <iostream> #include <stdio.h> #include <algorithm> #include <string.h> using namespace std; typedef long long LL; LL dp[30][3]; /** * dp[i][0],表示长度为i，不存在不吉利数字 * dp[i][1],表示长度为i, 不存在不吉利数字，且最高位为9 * dp[i][2],表示长度为i, 存在不吉利数字 */ void Init() { int i; dp[0][0] = 1; for(i=1; i<30; i++) { dp[i][0] = 10*dp[i-1][0] - dp[i-1][1]; dp[i][1] = dp[i-1][0]; dp[i][2] = 10*dp[i-1][2] + dp[i-1][1]; } } LL Slove(LL n) { int i, a[65]={0}, len=0, flag=0; LL n1 = n, ans=0; while(n1) { a[++len] = n1%10; n1 /= 10; } for(i=len; i>=1; i--) { ans += dp[i-1][2]*a[i]; if(flag)///高位已经出现49了，后面随意 ans += dp[i-1][0]*a[i]; else if(!flag && a[i]>4) ans += dp[i-1][1]; if(a[i+1]==4 && a[i]==9) flag = 1; } return ans; } int main() { int T; Init(); scanf("%d", &T); while(T--) { LL n; scanf("%I64d", &n); printf("%I64d ", Slove(n+1)); } return 0; }