简单的数位dp

数位dp的模板

hdu 2089

设dp[len][flag]，flag = 1表示前一位是6,否则前一位不是6.

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 50010;

int dig[10];
int dp[10][2];

int dfs(int len, int flag, int up)
{
    if(len == 0)
        return 1;
    if(!up && dp[len][flag] != -1)
        return dp[len][flag];
    int n = up == 1?dig[len]:9;
    int res = 0;
    for(int i = 0; i <= n; i++)
    {
        if(i == 4 || (flag == 1 && i == 2))
            continue;
        res += dfs(len-1,i==6?1:0,up&&(i==n) );
    }
    if(!up)
        dp[len][flag] = res;
    return res;
}

int cal(int num)
{
    int cnt = 0;
    while(num)
    {
        dig[++cnt] = num%10;
        num /= 10;
    }
    memset(dp,-1,sizeof(dp));
    return dfs(cnt,0,1);
}
int main()
{
    int a,b;
    while(~scanf("%d %d",&a,&b))
    {
        if(a == 0 && b == 0)
            break;
        printf("%d\n",cal(b) - cal(a-1) );
    }
    return 0;
}

hdu 3555

不能出现49，与上题类似

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 4010;

int dig[100];
LL dp[100][4];

LL dfs(int len, int flag, int up)
{
    if(len == 0)
        return flag == 2;
    if(!up && dp[len][flag] != -1)
        return dp[len][flag];

    int n = up ? dig[len] : 9;
    LL res = 0;

    for(int i = 0; i <= n; i++)
    {
        int tflag = flag;
        if(flag == 0 && i == 4)
            tflag = 1;
        else if(flag == 1 && i == 9)
            tflag = 2;
        else if(flag == 1 && i != 4)
            tflag = 0;
        res += dfs(len-1,tflag,up&&(i==n));
    }
    if(!up)
        dp[len][flag] = res;
    return res;
}

LL cal(LL num)
{
    int len = 0;
    while(num)
    {
        dig[++len] = num % 10;
        num /= 10;
    }
    memset(dp,-1,sizeof(dp));
    return dfs(len,0,1);
}

int main()
{
    int test;
    LL n;
    scanf("%d",&test);
    while(test--)
    {
        scanf("%I64d",&n);
        printf("%I64d\n",cal(n));
    }
    return 0;
}

SCOI2009

要求两个相邻的数之差大于等于2，对于第一位是特殊的。所以dfs的时候加一个变量first表示前面是否全是前导0。

#include <stdio.h>
#include <iostream>
#include <map>
#include <set>
#include <list>
#include <stack>
#include <vector>
#include <math.h>
#include <string.h>
#include <queue>
#include <string>
#include <stdlib.h>
#include <algorithm>
#define LL __int64
#define eps 1e-12
#define PI acos(-1.0)
using namespace std;
const int INF = 0x3f3f3f3f;
const int maxn = 4010;

int dig[15];
int dp[15][15]; //dp[i][j]表示第i位它的上一位即i+1位为j


int dfs(int len, int pre, int up, int first)
{
	if(len == 0)
		return first == 0;
	if(!up && dp[len][pre] != -1 && !first)
		return dp[len][pre];
	int n = up ? dig[len] : 9;
	int res = 0;
	for(int i = 0; i <= n; i++)
	{
		if(first) //当前这一位是第一位，当i==0时，第i-1为也是第一位，否则不是
			res += dfs(len-1,i,up&&i==n,first&&i==0);
		else if(abs(pre-i) >= 2)
			res += dfs(len-1,i,up&&i==n,first);
	}
	if(!up && !first)
		dp[len][pre] = res;
	return res;
}

int cal(int num)
{
	int len = 0;
	while(num)
	{
		dig[++len] = num%10;
		num /= 10;
	}
	memset(dp,-1,sizeof(dp));
	return dfs(len,0,1,1);
}

int main()
{
	int a,b;
	while(~scanf("%d %d",&a,&b))
	{
		printf("%d\n",cal(b) - cal(a-1));
	}
	return 0;

}