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  • codeforces 698A

    A. Vacations
    time limit per test
    1 second
    memory limit per test
    256 megabytes
    input
    standard input
    output
    standard output

    Vasya has n days of vacations! So he decided to improve his IT skills and do sport. Vasya knows the following information about each of this n days: whether that gym opened and whether a contest was carried out in the Internet on that day. For the i-th day there are four options:

    1. on this day the gym is closed and the contest is not carried out;
    2. on this day the gym is closed and the contest is carried out;
    3. on this day the gym is open and the contest is not carried out;
    4. on this day the gym is open and the contest is carried out.

    On each of days Vasya can either have a rest or write the contest (if it is carried out on this day), or do sport (if the gym is open on this day).

    Find the minimum number of days on which Vasya will have a rest (it means, he will not do sport and write the contest at the same time). The only limitation that Vasya has — he does not want to do the same activity on two consecutive days: it means, he will not do sport on two consecutive days, and write the contest on two consecutive days.

    Input

    The first line contains a positive integer n (1 ≤ n ≤ 100) — the number of days of Vasya's vacations.

    The second line contains the sequence of integers a1, a2, ..., an (0 ≤ ai ≤ 3) separated by space, where:

    • ai equals 0, if on the i-th day of vacations the gym is closed and the contest is not carried out;
    • ai equals 1, if on the i-th day of vacations the gym is closed, but the contest is carried out;
    • ai equals 2, if on the i-th day of vacations the gym is open and the contest is not carried out;
    • ai equals 3, if on the i-th day of vacations the gym is open and the contest is carried out.
    Output

    Print the minimum possible number of days on which Vasya will have a rest. Remember that Vasya refuses:

    • to do sport on any two consecutive days,
    • to write the contest on any two consecutive days.
    Examples
    input
    4
    1 3 2 0
    output
    2
    input
    7
    1 3 3 2 1 2 3
    output
    0
    input
    2
    2 2
    output
    1
    Note

    In the first test Vasya can write the contest on the day number 1 and do sport on the day number 3. Thus, he will have a rest for only 2 days.

    In the second test Vasya should write contests on days number 1, 3, 5 and 7, in other days do sport. Thus, he will not have a rest for a single day.

    In the third test Vasya can do sport either on a day number 1 or number 2. He can not do sport in two days, because it will be contrary to the his limitation. Thus, he will have a rest for only one day.

    题目的意思为0表示健身房和比赛都没有,1表示健身房开门,没有比赛,2表示健身房关门,有比赛,3表示健身房开门也有比赛。

    输入天数n,计算最少休息的天数,要求连续两天不能做同样的事情。

    使用DP的方法来做,有休息,健身和比赛三个状态,

    0表示那天休息,

    1表示健身,

    2表示比赛。

    因为要求最小值,所以dp数组的数初始化为一个充分大的数。

    先初始化第一组数据。

    dp[0][0] = 1;
    if (a[0] == 1) dp[0][1] = 0;
    if (a[0] == 2) dp[0][2] = 0;
    if (a[0] == 3) dp[0][2] = dp[0][1] = 0;

    然后更新每天的休息天数:

    dp[i][0] = min(min(dp[i - 1][0], dp[i - 1][1]), dp[i - 1][2]) + 1;
    dp[i][1] = min(min(dp[i - 1][0], dp[i - 1][2]), dp[i - 1][1] + 1);
    dp[i][2] = min(min(dp[i - 1][0], dp[i - 1][1]), dp[i - 1][2] + 1);

    #include<iostream>
    #include<algorithm>
    using namespace std;
    const int MAXN= 100;
    int a[MAXN], dp[MAXN][3];
    int main()
    {
        int n, i;
        cin >> n;
        for (i = 0;i < n;i++)
            cin >> a[i];
        memset(dp, MAXN, sizeof(dp));
        dp[0][0] = 1;
        if (a[0] == 1) dp[0][1] = 0;
        if (a[0] == 2) dp[0][2] = 0;
        if (a[0] == 3) dp[0][2] = dp[0][1] = 0;
        for (i = 1;i < n;i++)
        {
            dp[i][0] = min(min(dp[i - 1][0], dp[i - 1][1]), dp[i - 1][2]) + 1;
            if (a[i] == 1) dp[i][1] = min(min(dp[i - 1][0], dp[i - 1][2]), dp[i - 1][1] + 1);
            if (a[i] == 2) dp[i][2] = min(min(dp[i - 1][0], dp[i - 1][1]), dp[i - 1][2] + 1);
            if (a[i] == 3)
            {
                dp[i][1] = min(min(dp[i - 1][0], dp[i - 1][2]), dp[i - 1][1] + 1);
                dp[i][2] = min(min(dp[i - 1][0], dp[i - 1][1]), dp[i - 1][2] + 1);
            }
        }
        cout << min(min(dp[n - 1][0], dp[n - 1][1]), dp[n - 1][2]) << endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/chenruijiang/p/8310264.html
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