HDU 2089 不要62(数位DP，三种姿势)

ACM

题目地址：HDU 2089

题意：

中文题意，不解释。

分析：

1. 100w的数据，暴力打表能过
2. 先初始化dp数组，表示前i位的三种情况，再进行推算
3. 直接dfs，一遍搜一变记录，可能有不饥渴的全部算和饥渴的部分算情况，记录只能记录全部算（推荐看∑大的详细题解Orz)

代码：

1. 暴力 （以前写的）

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  File:        2089_bf.cpp
*  Create Date: 2014-03-31 20:28:46
*  Descripton:  brute force way
*/

#include <cstdio>
#define RII(x,y) scanf("%d%d",&x,&y)
#define PIN(x) printf("%d\n",x)
#define repf(i,a,b) for(int i=(a);i<=(b);i++)

const int N = 1000010;

int f[N], n, m;

bool check(int r) {
	while (r) {
		if (r % 10 == 4) return false;
		if (r % 100 == 62) return false;
		r /= 10;
	}
	return true;
}

void init() {
	repf (i, 1, 1000000)
		if (check(i))
			f[i] = f[i - 1] + 1;
		else
			f[i] = f[i - 1];
}

int main()
{
	init();
	while (RII(n, m) && (n || m)) {
		PIN(f[m] - f[n - 1]);
	}
	return 0;
}

2. DP_1

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  File:        2089.cpp
*  Create Date: 2014-07-26 09:55:48
*  Descripton:
*/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

#define RII(x,y) scanf("%d%d",&x,&y)
#define PIN(x) printf("%d\n",x)
#define repf(i,a,b) for(int i=(a);i<=(b);i++)
#define repd(i,a,b) for(int i=(a);i>=(b);i--)

const int N = 10;

int n, m;
int bits[N];
int dp[N][3];	// [len][x]
				// 0->luck
				// 1->luck and highest is 2
				// 2->unluck

void init() {
	dp[0][0] = 1;	// len is 0 and luck is 1, becauses the dp[1][1] will equal it.
	dp[0][1] = dp[0][1] = 0;

	repf(i, 1, N - 1) {

		dp[i][0] = dp[i - 1][0] * 9		// i-1_luck without 4
					- dp[i - 1][1];		// and without 62

		dp[i][1] = dp[i - 1][0];		// equal to i-1_luck_2 + '6'

		dp[i][2] = dp[i - 1][2] * 10	// unluck is always unluck
			+ dp[i - 1][0]				// i-1_luck + '4'
			+ dp[i - 1][1];				// i-1_luck_2 + '6'

	}
}

int solve(int num) {
	int len = 0, rec = num, ans, flag;

	// get bits array
	while (num) {
		bits[++len] = num % 10;
		num /= 10;
	}
	bits[len + 1] = 0;

	ans = 0;	// the unluck num
	flag = 0;	// if appear unluck
	repd (i, len, 1) {
		ans += dp[i - 1][2] * bits[i];	// unluck is always unluck

		if (flag) {						// if unluck appeared
			ans += dp[i - 1][0] * bits[i];	// all luck become unluck
		} else {
			if (bits[i] > 4) {
				ans += dp[i - 1][0];	// i-1_luck + '4'
			}
			if (bits[i] > 6) {
				ans += dp[i - 1][1];	// i-1_luck_2 + '6'
			}
			if (bits[i + 1] == 6 && bits[i] > 2) {
				ans += dp[i][1];
			}
		}

		if (bits[i] == 4 || (bits[i + 1] == 6 && bits[i] == 2)) {
			flag = 1;
		}
	}

	return rec - ans;
}

int main() {
	init();
	while (~RII(n, m) && (n || m)) {
		PIN(solve(m + 1) - solve(n));
	}
	return 0;
}

3. DP_2(DFS)

/*
*  Author:      illuz <iilluzen[at]gmail.com>
*  File:        2089_dfs.cpp
*  Create Date: 2014-07-26 15:21:12
*  Descripton:  dfs version
*/

#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
#define RII(x,y) scanf("%d%d",&x,&y)
#define PIN(x) printf("%d\n",x)

const int N = 10;

int bits[N], dp[N][2];	// dp[i][is6] is the number of i-len digits with (if prevent number is 6), its digits are from 0-9

// the rest length, Is the prevent digit 6, Is the digit max
int dfs(int len, bool is6, bool ismax) {
	if (len == 0)
		return 1;
	if (!ismax && dp[len][is6] >= 0)		// dp
		return dp[len][is6];

	int cnt = 0, mmax = (ismax? bits[len] : 9);
	for (int i = 0; i <= mmax; i++) {
		if (i == 4 || is6 && i == 2)	// unluck digit
			continue;
		cnt += dfs(len - 1, i == 6, ismax && i == mmax);
	}

	return ismax ? cnt : dp[len][is6] = cnt;	// remember the result
}

int solve(int num) {
	int len = 0;
	while (num) {
		bits[++len] = num % 10;
		num /= 10;
	}

	return dfs(len, false, true);
}

int main() {
	int n, m;
	memset(dp, -1, sizeof(dp));
	while (~scanf("%d%d", &n, &m) && (n || m)) {
		PIN(solve(m) - solve(n - 1));
	}
	return 0;
}

HDU 2089 不要62(数位DP，三种姿势),布布扣,bubuko.com