数位dp

1. Codeforces 55D Beautiful numbers

Problem : 询问一个区间内有多少个数，可以整除其每一位的数字。
**Solution : **记录一下前若干位数字的lcm和前缀和，由于1~9的lcm是2520，所以前缀和可以对2520取模。对于合法的lcm压缩一下状态，就可以存下来了。

#include <bits/stdc++.h>

using namespace std;

const int N = 20;
const int SUM = 2520;

int a[N];
int index[SUM];
long long dp[N][50][SUM];

int gcd(int x, int y)
{
	return y ? gcd(y, x % y) : x;
}

int lcm(int x, int y)
{
	return x * y / gcd(x, y);
}

long long dfs(int pos, int prelcm, int presum, int limit, int leed)
{
	if (pos == 0)
	{
		if (presum % prelcm == 0 && !leed) return 1;
		return 0;
	}
	if (!limit && !leed && ~dp[pos][index[prelcm]][presum]) return dp[pos][index[prelcm]][presum];
	int u = limit ? a[pos] : 9;
	long long ans = 0;
	for (int i = 0; i <= u; ++i)
	{
		ans += dfs(pos - 1, (i == 0) ? prelcm : lcm(prelcm, i), (presum * 10 + i) % SUM, 
					limit && i == u, leed && i == 0 );
	}
	if (!limit && !leed) dp[pos][index[prelcm]][presum] = ans;
	return ans;
}
long long solve(long long x)
{
	int len = 0;
	while (x)
	{
		a[++len] = x % 10;
		x /= 10;
	}
	return dfs(len, 1, 0, 1, 1);
}
int main()
{
	cin.sync_with_stdio(0);
	int tot;
	index[0] = tot = 1;
	for (int i = 1; i < SUM; ++i)
	{
		if (SUM % i == 0)
		{
			index[i] = ++tot;
		}
	}
	memset(dp, -1, sizeof(dp));
	int T; cin >> T;
	for (int i = 0; i < T; ++i)
	{
		long long l, r;
		cin >> l >> r;
		cout << solve(r) - solve(l - 1) << endl;
	}
}

2. HDU 4352 XHXJ'S LIS

**Problem : ** 询问一个区间内有多少个数 ，最长严格上升序列长度为k。
**Solution : ** 用类似于nlogn求最长不下降序列的思路，记录下每一位长度的最小数字。由于数字大小为0~9，且具有单调性质，直接用1<<10来表示每个数字出现过。在形成下一个状态时，找到大于等于当前数字的以为并替换。需要考虑如果有前导0并且当且数字是0的话是不记录状态的。

#include <iostream>
#include <cstring>

using namespace std;

long long dp[20][2048][11];
int a[20];

int get(int state)
{
	int len = 0;
	while (state)
	{
		if (state % 2 == 1) len++;
		state /= 2;
	}
	return len;
}

int next(int state, int num, int leed)
{
	if (leed && num == 0) return state;
	for (int i = num; i <= 9; ++i)
		if (state & (1 << i))
			return (state ^ (1 << i)) | (1 << num);
	return state | (1 << num);
}

long long dfs(int pos, int k, int state = 0, int limit = 1, int leed = 1)
{
	if (!limit && !leed && ~dp[pos][state][k]) return dp[pos][state][k];
	if (pos == 0)
	{
		return get(state) == k;
	}	
	int u = limit ? a[pos] : 9;
	long long ans = 0;
	for (int i = 0; i <= u; ++i)
	{
		ans += dfs(pos - 1, k, next(state, i, leed), limit && i == u, leed && i == 0);
	}
	if (!limit && !leed) dp[pos][state][k] = ans;
	return ans;
}

long long solve(long long x, int k)
{
	int len = 0;
	while (x)
	{
		a[++len] = x % 10;
		x /= 10;
	}
	return dfs(len, k);
}	

int main()
{
	memset(dp, -1, sizeof(dp));
	cin.sync_with_stdio(0);
	int T; cin >> T;
	for (int cas = 1; cas <= T; ++cas)
	{
		long long l, r;
		int k;
		cin >> l >> r >> k;
		cout << "Case #" << cas << ": " << solve(r, k) - solve(l - 1, k) << endl; 
	}
	/*for (int i = 123; i <= 321; ++i)
	{
		cout << i << " " << solve(i, 2) - solve(i - 1, 2) << endl;
	}*/
}

3. HDU 4507 吉哥系列故事――恨7不成妻

**Problem : **询问一个区间里满足以下性质的数字的平方和：没有一位是7， 数字和不是7的倍数， 这个数不是7的倍数
**Solution : **用三个状态是否出现过7，数字之和，数位之和就可以判断是否是合法的数字。询问要求的是平方和，根据平方和公式(10a+b)^2可以得到只需记录平方和，和以及数量就可以递推得到下一位的平方和。

#include <bits/stdc++.h>
using namespace std;

const int N = 20;
const int mo =1e9 + 7;
struct node
{
	long long num;
	long long sum;
	long long squre;
	node(long long num_ = 0, long long sum_ = 0, long long squre_ = 0):num(num_), sum(sum_), squre(squre_){}
};
node dp[N][2][7][7];
long long power[N];
int a[N];
void upd(long long &x, long long y)
{
	x = (x + y) % mo;
}
node dfs(int pos, int y7, int digit, int presum, int limit)
{
	if (pos == 0)
	{
		if (!y7 && digit && presum) return node(1, 0, 0); else return node(0, 0, 0);
	}
	if (!limit && ~dp[pos][y7][digit][presum].num) return dp[pos][y7][digit][presum];
	int u = limit ? a[pos] : 9;
	node ans;
	for (int i = 0; i <= u; ++i)
	{
		node nxt = dfs(pos - 1, y7 || i == 7, (digit + i) % 7, (presum * 10 + i) % 7, limit && i == u);
		upd(ans.num, nxt.num);	
		upd(ans.sum, (nxt.sum + power[pos - 1] * i % mo * nxt.num) % mo);
	   	upd(ans.squre, power[pos - 1] * power[pos - 1]  % mo * i * i % mo * nxt.num % mo);
		upd(ans.squre, power[pos - 1] * 2 * i % mo * nxt.sum % mo);
		upd(ans.squre, nxt.squre);
	}
	if (!limit) dp[pos][y7][digit][presum] = ans;
	return ans;
}

long long solve(long long x)
{
	int len = 0;
	while (x)
	{
		a[++len] = x % 10;
		x /= 10;
	}
	return dfs(len, 0, 0, 0, 1).squre;
}

int main()
{
	power[0] = 1;
	for (int i = 1; i < N; ++i) power[i] = power[i - 1] * 10 % mo;
	cin.sync_with_stdio(0);
	int T; cin >> T;
	memset(dp, -1, sizeof(dp));
	for (int cas = 0; cas < T; ++cas)
	{
		long long l, r; cin >> l >> r;
		cout << ((solve(r) - solve(l - 1)) % mo + mo) % mo << endl;
	}
	return 0;
}

4. SPOJ BALNUM Balanced Numbers

**Problem: ** 询问一个区间内有多少个数字满足：奇数数字出现偶数次，偶数数字出现奇数次。
**Solution: ** 用一个三进制的数记录下每个数字出现的状态，0，1，2分别表示奇数次，偶数次和没出现过。

#include <bits/stdc++.h>

using namespace std;

const int N = 22;
const int M = 60000;

typedef unsigned long long ull;

ull dp[N][M];
int a[N];
int power[N];

int change(int status, int num)
{
	int tot = status;
	for (int i = 1; i <= num; i++) tot /= 3;
	switch (tot % 3)
	{
	case 0:return status + power[num];break;
	case 1:return status + power[num];break;
	case 2:return status - power[num];break;
	}
}

ull dfs(int pos, int status, int limit, int leed)
{
	if (pos == 0)
	{
		//cout << pos << " " << status << endl;
		int tot = status;
		for (int i = 0; i <= 9; ++i)
		{
			switch (tot % 3)
			{
			case 0:break;
			case 1:if (i % 2 == 1) return 0;break;
			case 2:if (i % 2 == 0) return 0;break;
			}
			tot /= 3;
		}	
		return 1;
	}
	if (!limit && !leed && ~dp[pos][status]) return dp[pos][status];
	int u = limit ? a[pos] : 9;
	ull ans = 0;
	for (int i = 0; i <= u; ++i)
	{
		ans += dfs(pos - 1, (leed && i == 0) ? status : change(status,i), limit && i == u, leed && i == 0);
	}
	if (!limit && !leed) dp[pos][status] = ans;
	return ans;
}
ull solve(ull x)
{
	int len = 0;
	while (x)
	{
		a[++len] = x % 10;
		x /= 10;
	}
	return dfs(len, 0, 1, 1);
}
int main()
{
	power[0] = 1;
	for (int i = 1; i <= 9; ++i) power[i] = power[i - 1] * 3;
	cin.sync_with_stdio(0);
	int T; cin >> T;
	memset(dp, -1, sizeof(dp));
	for (int i = 1; i <= T; i++)
	{
		ull l, r;
		cin >> l >> r;
		cout << solve(r) - solve(l - 1) << endl;
	}	
	return 0;
}

5. HDU 4734 F(x)

**Problem: **定义F(x) = An * 2^(n-1) + ... + A 2 * 2 + A 1 * 1.询问0~B内有多少个数满足f(x) <= f(A)。 10000组数据。
**Solution: **由于有多组数据，因此需要考虑记录下与输入无关的状态，来优化程序速度。倒着来存储状态，记录下当前数值距离0还差多少，从而与A无关。

#include <bits/stdc++.h>

using namespace std;

const int N = 12;
const int M = 10008;

int dp[N][M];
int a[N];

int dfs(int pos, int presum, int limit)
{
	if (presum < 0) return 0;
	if (pos == 0) return 1;
	if (!limit && ~dp[pos][presum]) return dp[pos][presum];
	int u = limit ? a[pos] : 9;
	int ans = 0;
	for (int i = 0; i <= u; ++i)
	{
		ans += dfs(pos - 1, presum - (1 << pos - 1) * i, limit && i == u);
	}
	if (!limit) dp[pos][presum] = ans;
	return ans;
}

int solve(int A, int B)
{
	int len = 0, sum = 0;
	while (A)
	{
		a[++len] = A % 10;
		A /= 10;
	}
	for (int i = 1; i <= len; ++i) sum = sum + a[i] * (1 << i - 1);
	len = 0;
	while (B)
	{
		a[++len] = B % 10;
		B /= 10;
	}
	return dfs(len, sum, 1);
}

int main()
{
	cin.sync_with_stdio(0);
	int T; cin >> T;
	memset(dp, -1 ,sizeof(dp));
	for (int i = 1; i <= T; ++i)
	{
		int A, B; cin >> A >> B;
		cout << "Case #" << i << ": " << solve(A, B) << endl; 
	}
	return 0;
}