POJ 1008 简单模拟题

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e.... 虽然这是一道灰常简单的模拟题、但是米做的时候没有读懂第二个日历的计时方法。然后捏。敲完之后华丽的WA了进一个点。坑点就在一年的最后一天你是该输出本年的。e ...但是我好想并没有。、看discuss里好想被坑的人还不少。总天数能直接整除260的时候。年数要减1。

不喜欢做模拟.....5555....

附代码：

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


string Maya_name[20] = {"pop", "no", "zip", "zotz", "tzec",  "xul", "yoxkin", "mol", "chen", "yax",  "zac", "ceh", "mac", "kankin", "muan",  "pax", "koyab", "cumhu", "uayet"};
string Holly_name[21] = {"imix", "ik", "akbal", "kan", "chicchan",   "cimi", "manik", "lamat", "muluk", "ok",   "chuen", "eb", "ben", "ix", "mem",   "cib", "caban", "eznab", "canac", "ahau"};

int main()
{
    map<string, int>m;
    m.clear();
    for (int i=0; i<19; ++i)
    {
        m[Maya_name[i]] = i;
    }

    map<int, string>mm;
    mm.clear();
    for (int i=0; i<20; ++i)
    {
        mm[i+1] = Holly_name[i];
    }

    int t, total;
    int day, year;
    char temp;
    string month;
    while(cin >> t)
    {
        cout << t << endl;
        while(t--)
        {
            total = 0;
            cin >> day >> temp >> month >> year;
            int tm = m[month];
            total += year * 365 + tm * 20 + day + 1;
           // cout << total << "==\n";
            year = total / 260;
            int tot = total;
            tot %= 260;
            tm = tot % 20;
            if (tm == 0) tm = 20;
            day = total % 13;
            if (day == 0) day = 13;
            if (total % 260 == 0)
            {
                year -= 1;
            }
            cout << day << ‘ ‘ << mm[tm] << ‘ ‘ << year << endl;
        }
    }
    return 0;
}

POJ 1008 简单模拟题

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原文地址：http://www.cnblogs.com/icode-girl/p/4600776.html