HDU 2089 HDU3555 数位DP

HUD 2089

不要62

Time Limit: 1000/1000 MS (Java/Others)    Memory Limit: 32768/32768 K (Java/Others)
Total Submission(s): 43142    Accepted Submission(s): 15815

Problem Description

杭州人称那些傻乎乎粘嗒嗒的人为62（音：laoer）。
杭州交通管理局经常会扩充一些的士车牌照，新近出来一个好消息，以后上牌照，不再含有不吉利的数字了，这样一来，就可以消除个别的士司机和乘客的心理障碍，更安全地服务大众。
不吉利的数字为所有含有4或62的号码。例如：
62315 73418 88914
都属于不吉利号码。但是，61152虽然含有6和2，但不是62连号，所以不属于不吉利数字之列。
你的任务是，对于每次给出的一个牌照区间号，推断出交管局今次又要实际上给多少辆新的士车上牌照了。

Input

输入的都是整数对n、m（0<n≤m<1000000），如果遇到都是0的整数对，则输入结束。

Output

对于每个整数对，输出一个不含有不吉利数字的统计个数，该数值占一行位置。

Sample Input

1 100 0 0

Sample Output

80

Author

qianneng

做法一：

#include <stdio.h>
#include <string.h>
#include <iostream>
using namespace std;

int dp[10][3];

void init()
{
    int len = 6;
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for(int i = 1; i <= len; i++)
    {
        dp[i][0] = dp[i-1][0] * 9 - dp[i-1][1]; //首位不放4 不在次位是2的情况下放6
        dp[i][1] = dp[i-1][0];//首位放的是2
        dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1] + dp[i-1][0]; //在2前面放6、放4 或者 在不合格的情况下随便放
    }
}

int solve(int x)
{
    int a[10],n = x, ans = 0;
    int len = 0;
    int flag = 0;
    while (x)
    {
        a[++len] = x%10;
        x /= 10;
    }
    a[len + 1] = 0;
    for(int i = len; i >= 1; i--)
    {
        ans += dp[i-1][2] * a[i];
        if(flag)
        {
            ans += dp[i-1][0] * a[i];
        }
        if(!flag && a[i] > 4)
        {
            ans += dp[i-1][0];
        }
        if(!flag && a[i+1] == 6 && a[i] > 2)
        {
            ans += dp[i][1];
        }
        if(!flag && a[i] > 6)
        {
            ans += dp[i-1][1];
        }
        if(a[i] == 4 || (a[i] == 2 && a[i+1] == 6))
        {
            flag = 1;
        }
    }
    return n - ans;
}

int main()
{
    int l, r;
    init();
    while(scanf("%d%d", &l, &r), l || r)
    {
        printf("%d
", solve(r + 1) - solve(l));
    }
    return 0;
}

做法二（记忆化搜索）：

#include <iostream>
#include <string.h>
#include <stdio.h>
using namespace std;

int a[10];
int dp[10][2];

int dfs(int pre, int pos, int status, bool limit)
{
    if(pos == 0) return 1;
    if(!limit && dp[pos][status] != -1) return dp[pos][status];
    int up = limit ? a[pos] : 9;
    int ans = 0;
    for(int i = 0; i <= up; i++)
    {
        if(i == 4 ) continue;
        if(pre == 6 && i == 2) continue;
        ans += dfs(i, pos-1, i == 6, limit && i == up);
    }
    if(!limit) dp[pos][status] = ans;
    return ans;
}

int solve(int x)
{
    int len = 0;
    memset(dp, -1, sizeof(dp));
    while(x)
    {
        a[++len] = x%10;
        x /= 10;
    }
    return dfs(-1, len, 0, true);
}

int main()
{
    int l, r;
    while(scanf("%d%d", &l, &r), l || r)
    {
        printf("%d
", solve(r) - solve(l - 1));
    }
    return 0;
}

有了上面的思路，可以解决其他类似的问题。

HDU 3555 试试水

Bomb

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 131072/65536 K (Java/Others)
Total Submission(s): 18374    Accepted Submission(s): 6781

Problem Description

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?

Input

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.

Output

For each test case, output an integer indicating the final points of the power.

Sample Input

3 1 50 500

Sample Output

0 1 15

Hint

From 1 to 500, the numbers that include the sub-sequence "49" are "49","149","249","349","449","490","491","492","493","494","495","496","497","498","499", so the answer is 15.

Author

fatboy_cw@WHU

做法一：

#include <iostream>
#include <stdio.h>
#include <string.h>
using namespace std;
typedef long long LL;

LL dp[30][3];
int a[30];

void init()
{
    memset(dp, 0, sizeof(dp));
    dp[0][0] = 1;
    for(int i = 1; i <= 25; i++)
    {
        dp[i][0] = dp[i-1][0] * 10 - dp[i-1][1]; //减去首位是6 次位是2
        dp[i][1] = dp[i-1][0];        //首位是2
        dp[i][2] = dp[i-1][2] * 10 + dp[i-1][1];//加上首位是6 次位是2
    }
}

LL solve(LL x)
{
    int len = 0;
    while(x)
    {
        a[++len] = x%10;
        x /= 10;
    }
    a[len + 1] = 0;
    LL ans = 0LL;
    bool flag = false;
    for(int i = len; i >= 1; i--)
    {
        ans += dp[i-1][2] * a[i];
        if(flag)
            ans += dp[i-1][0] * a[i];
        if(!flag && a[i] > 4)
            ans += dp[i-1][1];
        if(a[i+1] == 4 && a[i] == 9)
            flag = true;
    }
    return ans;
}

int main()
{
    init();
    int q;
    LL n;
    scanf("%d", &q);
    while(q--)
    {
        scanf("%lld
", &n);
        printf("%lld
", solve(n+1));
    }
    return 0;
}

做法二（记忆化搜索）：

#include <iostream>
#include <string.h>
#include <stdio.h>
typedef long long LL;
int a[30];
LL dp[30][3];

LL dfs(int pre, int pos, int status, bool limit)
{
    if(pos == 0) return status == 2;
    if(!limit && dp[pos][status] != -1)
    {
        return dp[pos][status];
    }
    LL ans = 0LL;
    int up = limit ? a[pos] : 9;
    for(int i = 0; i <= up; i++)
    {
        int nstatus;
        if(status == 2 || pre == 4 && i == 9)
            nstatus = 2;
        else if(i == 4)
            nstatus = 1;
        else
            nstatus = 0;
        ans += dfs(i, pos-1, nstatus, limit && i == up);
    }
    if(!limit) dp[pos][status] = ans;
    return ans;
}

LL solve(LL x)
{
    LL n = x, len = 0;
    memset(dp, -1, sizeof(dp));
    while(x)
    {
        a[++len] = x%10;
        x /= 10;
    }
    a[len + 1] = 0;
    return dfs(-1, len, 0, true);
}


int main()
{
    LL n;
    int q;
    scanf("%d", &q);
    while(q--)
    {
        scanf("%lld", &n);
        printf("%lld
",solve(n));
    }
    return 0;
}