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  • 【codeforces 760A】Petr and a calendar

    time limit per test2 seconds
    memory limit per test256 megabytes
    inputstandard input
    outputstandard output
    Petr wants to make a calendar for current month. For this purpose he draws a table in which columns correspond to weeks (a week is seven consequent days from Monday to Sunday), rows correspond to weekdays, and cells contain dates. For example, a calendar for January 2017 should look like on the picture:

    Petr wants to know how many columns his table should have given the month and the weekday of the first date of that month? Assume that the year is non-leap.

    Input
    The only line contain two integers m and d (1 ≤ m ≤ 12, 1 ≤ d ≤ 7) — the number of month (January is the first month, December is the twelfth) and the weekday of the first date of this month (1 is Monday, 7 is Sunday).

    Output
    Print single integer: the number of columns the table should have.

    Examples
    input
    1 7
    output
    6
    input
    1 1
    output
    5
    input
    11 6
    output
    5
    Note
    The first example corresponds to the January 2017 shown on the picture in the statements.

    In the second example 1-st January is Monday, so the whole month fits into 5 columns.

    In the third example 1-st November is Saturday and 5 columns is enough.

    【题目链接】:http://codeforces.com/contest/760/problem/A

    【题解】

    1,3,5,7,8,10,12是31天.其他的除了2月28天外都是30天.
    模拟现在到了第几天就好;
    给的是第一天的日期.
    所以从第一排到最后一排模拟一下就好.

    【完整代码】

    #include <bits/stdc++.h>
    
    using namespace std;
    
    const int days[14] = {0,31,28,31,30,31,30,31,31,30,31,30,31};
    
    int m,d;
    
    int main()
    {
        cin >> m >> d;
        int now = d,cnt = 1,c = 1;
        while (cnt<days[m])
        {
            cnt++;
            now++;
            if (now>7)
            {
                now = 1;
                c++;
            }
        }
        cout << c << endl;
        return 0;
    }
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  • 原文地址:https://www.cnblogs.com/AWCXV/p/7626708.html
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