HDU-3555 Bomb (数位dp 入门题)

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 20547    Accepted Submission(s): 7696

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

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.

#include 
using namespace std;
long long dp[29][10][2];
void init(){
	memset(dp, 0, sizeof(dp));
	for(int i = 0; i <= 9; ++i){
		dp[1][i][0] = 1;
	}
	for(int i = 2; i <= 19; ++i){
		for(int j = 0; j <= 9; ++j){
			for(int k = 0; k <= 9; ++k){
				if(j == 4 && k == 9){
					dp[i][j][1] += dp[i - 1][k][0];
					dp[i][j][1] += dp[i - 1][k][1];
				}
				else{
					dp[i][j][0] += dp[i - 1][k][0];
					dp[i][j][1] += dp[i - 1][k][1];
				}
			}
		}
	}
}
int main(){
	init();
	int T;
	long long x;
	scanf("%d", &T);
	while(T--){
		cin >> x;
		x++;
		vectors;
		while(x){
			s.push_back(x % 10);
			x /= 10;
		}
		int n = s.size() - 1;
		long long ans = 0;
		s.push_back(0);
		int flag = 0;
		for(int i = n; i >= 0; --i){
			for(int j = 0; j < s[i]; ++j){
				if(flag == 0)  ans += dp[i + 1][j][1];
				else ans += dp[i + 1][j][0] + dp[i + 1][j][1];
			}
			if(s[i] == 9 && s[i + 1] == 4) flag = 1;
		}
		printf("%I64d\n", ans);
	}
	return 0;
}

/*
题意：
1e18个数，问有多少个数的数位中包含连续的49。

思路：
数位dp，dp[i][j][0]表示最高位为第i位，且最高位为j时，数位中不包含49有多少种情况，
dp[i][j][1]表示最高位为第i位，且最高位为j时，数位中包含49有多少种情况。转移就很容
易写出来了。统计答案时注意如果已经出现49了，就不可以只统计包含49的情况。
*/