HDU 3555 Bomb 数位dp

题目链接：

http://acm.hdu.edu.cn/showproblem.php?pid=3555

Bomb

Time Limit: 2000/1000 MS (Java/Others)
Memory Limit: 131072/65536 K (Java/Others) #### 问题描述 > 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? #### 输入 > 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. #### 输出 > For each test case, output an integer indicating the final points of the power. #### 样例 > **sample input** > 3 > 1 > 50 > 500 > > **sample output** > 0 > 1 > 15

题意

给你一个数n，求1到n里面有多少个数其中存在子串49的。

题解

数位dp。

代码

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

const int maxn = 20;
typedef long long LL;

//dp[len][0]表示长度<=len,且不包含49的总数
//dp[len][1]表示长度<=len,且以9结尾但不包含49的总数
//dp[len][2]表示长度<=len,且包好49的总数
//dp[len][0]+dp[len][2]=所有长度小于等于len的数。
//dp[len][1]是dp[len][0]的一个子集。
LL dp[maxn][3];
void pre() {
	memset(dp, 0, sizeof(dp));
	dp[0][0] = 1;
	for (int i = 1; i < maxn; i++) {
		dp[i][0] = dp[i - 1][0] * 10 - dp[i - 1][1];
		dp[i][1] = dp[i - 1][0];
		dp[i][2] = dp[i - 1][2] * 10 + dp[i - 1][1];
	}
}

int main() {
	pre();
	int tc;
	scanf("%d", &tc);
	while (tc--) {
		LL x; scanf("%I64d", &x);
		LL ans = 0;
		int arr[maxn], tot=0;
		while (x) { arr[++tot]=x%10; x /= 10; }
		bool flag = false; //标记x的高位时否已经出现49的组合。
		for (int i = tot; i >= 1; i--) {
			//这时考虑的是高位已经固定，这一位数为0（是可以有0的，因为高位还会有数，虽然最高位后面每数了，但它统计的是<=len,
			//而不是==len!!!）到arr[i]-1的所有情况中包好49的数
			ans += dp[i-1][2] * arr[i];
			if (flag) ans += dp[i-1][0] * arr[i];
			else if (arr[i] > 4) {//如果4刚好是arr[i]，那也是不能乱搞的！边界情况我们会一直往后推来考虑
				ans += dp[i - 1][1];
			}
			if (i + 1 <= tot&&arr[i] == 9 && arr[i + 1] == 4) {
				flag = 1;
			}
		}
		if (flag) ans++;
		printf("%I64d\n", ans);
	}
	return 0;
}

再来一发dfs:

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

const int maxn = 20;
typedef __int64 LL;

//dp[len][0]是给第len+1位不为4的时候用的
//dp[len][1]是给第len+1位为4的时候用的(有点像私人订制)
//我们只保存没有限制的状态，对于有限制的状态每次都要算，不过没有限制的状态出现的要不有限制的多的多。
LL dp[maxn][2],ten[maxn];
LL n;

int arr[maxn];
LL dfs(int len, int is4, int ismax) {
	if (len == 0) return 0;
	if (!ismax&&dp[len][is4] >= 0) return dp[len][is4];
	LL res = 0;
	int ed = ismax ? arr[len] : 9;
	for (int i = 0; i <= ed; i++) {
		if (i == 9 && is4) {
			//这里是可以直接算的哦
			res += ismax ? (n%ten[len - 1] + 1) : ten[len - 1];
		}
		else {
			res += dfs(len - 1, i == 4, ismax&&i == ed);
		}
	}
	return ismax ? res : dp[len][is4] = res;
}

LL solve(LL x) {
	int tot = 0;
	while (x) { arr[++tot] = x % 10; x /= 10; }
	return dfs(tot, false, true);
}

void init() {
	memset(dp, -1, sizeof(dp));
	ten[0] = 1;
	for (int i = 1; i < maxn; i++) ten[i] = ten[i - 1] * 10;
}

int main() {
	int tc;
	scanf("%d", &tc);
	init();
	while (tc--) {
		scanf("%I64d", &n);
		printf("%I64d\n", solve(n));
	}
	return 0;
}