CDOJ 1137 邱老师选妹子 数位dp

经典的不要62和4的题。。。

先附代码吧，，有时间再贴贴思路。。。反正网上一大堆（那我当时为什么只去盯着学校的dfs盯了那么久。。。。看用迭代写的，感觉容易理解多了。。。）

代码：

#include 
#include 
#include 
#include 
using namespace std;
#define maxn 1000005
int dp[8][12];
int lnum[10], rnum[10];
int l, r, llen, rlen, lans, rans;
void init()
{
	dp[0][0] = 1;
	for (int i = 1; i <= 7; ++i)
	{
		for (int j = 0; j <= 9; ++j)
		{
			for (int k = 0; k <= 9; ++k)
			{
				if (j != 4 && !(j == 6 && k == 2))
					dp[i][j] += dp[i - 1][k];
			}
		}
	}
}
void deal_num()
{
	int x = l;
	llen = 0, rlen = 0;
	memset(lnum, 0, sizeof(lnum));
	memset(rnum, 0, sizeof(rnum));
	while (x != 0)
	{
		lnum[++llen] = x % 10;
		x /= 10;
	}
	x = r;
	while (x != 0)
	{
		rnum[++rlen] = x % 10;
		x /= 10;
	}
}
void compute()
{
	lans = rans = 0;
	for (int i = llen; i >= 1; --i)
	{
		for (int j = 0; j < lnum[i]; ++j)
		{
			if (j != 4 && !(j == 2 && lnum[i + 1] == 6))
				lans += dp[i][j];
		}
		if (lnum[i] == 4 || (lnum[i] == 2 && lnum[i + 1] == 6))
			break;
	}
	for (int i = rlen; i >= 1; --i)
	{
		for (int j = 0; j < rnum[i]; ++j)
		{
			if (j != 4 && !(j == 2 && rnum[i + 1] == 6))
				rans += dp[i][j];
		}
		if (rnum[i] == 4 || (rnum[i] == 2 && rnum[i + 1] == 6))
			break;
	}
}
int main()
{
	//freopen("input.txt", "r", stdin);
	init();
	/*for (int i = 1; i <= 6; ++i)
	{
		for (int j = 0; j <= 9; ++j)
			printf("%d ", dp[i][j]);
		printf("\n");
	}*/
	while (scanf("%d%d", &l, &r) != EOF)
	{
		if (l == 0 && r == 0)
			break;
		++r;
		deal_num();
		compute();
		int ans = rans - lans;
		//printf("%d %d\n", lans, rans);
		printf("%d\n", ans);
	}
	//system("pause");
	//while (1);
	return 0;
}