我想我的大脑只是抛出一个内存异常并且崩溃了，
我的问题是我有一个SYSTEMTIME大小为3的类成员数组，它是用户定义的（从.lua中读取）
SYSTEMTIME m_MatchTime[3];
然后从文件中以这种方式阅读：
m_MatchTime[0].wDayOfWeek = static_cast<WORD>( m_Lua.GetGlobalNumber( "FirstDay" ) );
m_MatchTime[0].wHour = static_cast<WORD>( m_Lua.GetGlobalNumber( "FirstHour" ) );
m_MatchTime[0].wMinute = static_cast<WORD>( m_Lua.GetGlobalNumber( "FirstMinute" ) );
m_MatchTime[1].wDayOfWeek = static_cast<WORD>( m_Lua.GetGlobalNumber( "SecondDay" ) );
m_MatchTime[1].wHour = static_cast<WORD>( m_Lua.GetGlobalNumber( "SecondHour" ) );
m_MatchTime[1].wMinute = static_cast<WORD>( m_Lua.GetGlobalNumber( "SecondMinute" ) );
m_MatchTime[2].wDayOfWeek = static_cast<WORD>( m_Lua.GetGlobalNumber( "ThirdDay" ) );
m_MatchTime[2].wHour = static_cast<WORD>( m_Lua.GetGlobalNumber( "ThirdHour" ) );
m_MatchTime[2].wMinute = static_cast<WORD>( m_Lua.GetGlobalNumber( "ThirdMinute" ) );
现在我有一个方法：
SYSTEMTIME cTime;
GetLocalTime( &cTime );
我必须计算从三个用户定义的时间开始到更接近当前时间，然后计算剩余时间， （请注意星期日= 0，星期六= 6，还要注意只有wDayOfWeek，wHour和wMinute必须进行比较才能到达最近的地方）
编辑：现在我正在为 500bounty 寻找解决方案，请注意我想要的示例，
今天：第4天，第3小时，分钟0，
日期：第5天，第5小时，30分钟
到目前为止的剩余时间是：1天，2小时30分钟。
答案 0 :(得分：5)
鉴于问题域，似乎没有必要（甚至可取）强制执行严格的时间排序，您只想找到最接近给定标记值的一组时间。这将需要线性复杂性，但很容易实现。
我建议计算一个已知世纪的时差，在这种情况下，周日00:00:00（秒），然后比较每个时间点与该点之间的差异，看看哪个最接近。
#include <Windows.h>
#include <algorithm>
#include <iostream>
long seconds_from_sunday_epoch(const SYSTEMTIME& t)
{
size_t seconds = t.wDayOfWeek * 86400;
seconds += t.wHour * 3600;
seconds += t.wMinute * 60;
return seconds;
}
size_t timediff_2(const SYSTEMTIME& t0, const SYSTEMTIME& t1)
{
size_t seconds_diff = std::abs(
seconds_from_sunday_epoch(t0) -
seconds_from_sunday_epoch(t1));
return seconds_diff;
}
int main()
{
SYSTEMTIME m_MatchTime[3];
// Monday: 00:00
m_MatchTime[0].wDayOfWeek = 1;
m_MatchTime[0].wHour = 0;
m_MatchTime[0].wMinute = 0;
// Sunday: 01:00
m_MatchTime[1].wDayOfWeek = 0;
m_MatchTime[1].wHour = 1;
m_MatchTime[1].wMinute = 0;
// Wednesday: 15:30
m_MatchTime[2].wDayOfWeek = 3;
m_MatchTime[2].wHour = 15;
m_MatchTime[2].wMinute = 30;
// Sunday 23:00
SYSTEMTIME cTime;
cTime.wDayOfWeek = 0;
cTime.wHour = 23;
cTime.wMinute = 0;
std::cout << timediff_2(cTime, m_MatchTime[0]) << "\n";
std::cout << timediff_2(cTime, m_MatchTime[1]) << "\n";
std::cout << timediff_2(cTime, m_MatchTime[2]) << "\n";
}
答案 1 :(得分：3)
所以问题是你坐在一个圆圈上想要知道从d1到d2的右边（保持在同一周）或左边（一个值到下一个星期天）的距离更短。
首先，您应该将日期转换为公式分钟+小时* 60 +工作日* 60 * 24的值。这将给你一周的分钟。
#include <stdlib.h>
int minOfWeek (int d, int h, int m) {
return d*60*24+h*60+m;
}
接下来找到最小距离：
const int minutesInWeek=60*24*7;
int bestDistance (int minutes1, int minutes2) {
int d=abs (minutes1-minutes2);
int dNext=minutesInWeek-d;
return d<dNext?d:dNext;
}
因此，根据你的实际时间计算minOfWeek，用你所有的3次喂它到最佳距离并取最小的数字......
答案 2 :(得分：2)
标准C ++库让您可以通过将比较日期的“魔力”移动到仿函数，并使用带有自定义比较器的std::sort重载来优雅地解决这个问题。
以下是使用极少行代码（link to a quick test on ideone）执行此操作的方法：
class ClosestTo {
int minute_now;
int abs_minute(const SYSTEMTIME& t) const {
return 60 * (24 * t.wDayOfWeek + t.wHour) + t.wMinute;
}
int diff_to_now(const SYSTEMTIME& t) const {
int res = abs_minute(t) - minute_now;
// Has the date passed this week?
if (res < 0) {
// Yes, the date has passed - move to next week:
res += 7*24*60;
}
return res;
}
public:
ClosestTo(const SYSTEMTIME& now)
: minute_now(abs_minute(now)) {
}
// This is the operator the std::sort is going to call to determine ordering
bool operator() (const SYSTEMTIME& lhs, const SYSTEMTIME& rhs) const {
// Pick the date implying the shortest difference to minute_now
return diff_to_now(lhs) < diff_to_now(rhs);
}
};
就是这样，真的！有了这个比较器，您可以按如下方式对三个日期进行排序：
ClosestTo cmp(cTime);
sort(m_MatchTime, m_MatchTime+3, cmp);
现在最近的日期是零指数：
SYSTEMTIME &nearest = m_MatchTime[0];
答案 3 :(得分：0)
我已经找到了解决方案的算法，我知道它远非最专业的方式，但到目前为止它完美无瑕。
int main()
{
SYSTEMTIME m_MatchTime[3];
// Monday: 00:00
m_MatchTime[0].wDayOfWeek = 1;
m_MatchTime[0].wHour = 22;
m_MatchTime[0].wMinute = 4;
// Sunday: 01:00
m_MatchTime[1].wDayOfWeek = 4;
m_MatchTime[1].wHour = 1;
m_MatchTime[1].wMinute = 0;
// Wednesday: 15:30
m_MatchTime[2].wDayOfWeek = 6;
m_MatchTime[2].wHour = 15;
m_MatchTime[2].wMinute = 30;
// Sunday 23:00
SYSTEMTIME cTime;
cTime.wDayOfWeek = 3;
cTime.wHour = 14;
cTime.wMinute = 5;
/* std::cout << timediff_2(cTime, m_MatchTime[0]) << "\n";
std::cout << timediff_2(cTime, m_MatchTime[1]) << "\n";
std::cout << timediff_2(cTime, m_MatchTime[2]) << "\n";*/
vector<size_t>m_Time;
if( cTime.wDayOfWeek == 0 )
{
for( int i =0; i<3; i++ )
{
if( cTime.wDayOfWeek >= m_MatchTime[i].wDayOfWeek )
m_Time.push_back( timediff_2(cTime, m_MatchTime[i]) );
}
if( m_Time.size() == 0 ) //trim right
{
for( int i =0; i<3; i++ )
{
if( cTime.wDayOfWeek <= m_MatchTime[i].wDayOfWeek )
m_Time.push_back( timediff_2(cTime, m_MatchTime[i]) );
}
}
}
else
{
for( int i =0; i<3; i++ )
{
if( cTime.wDayOfWeek <= m_MatchTime[i].wDayOfWeek )
m_Time.push_back( timediff_2(cTime, m_MatchTime[i]) );
}
if( m_Time.size() == 0 ) //trim right
{
for( int i =0; i<3; i++ )
{
if( cTime.wDayOfWeek >= m_MatchTime[i].wDayOfWeek )
m_Time.push_back( timediff_2(cTime, m_MatchTime[i]) );
}
}
}
std::sort( m_Time.begin(), m_Time.end() );
SYSTEMTIME nearest;
if( m_Time.size() > 0 )
{
for( int l=0; l<3; l++ )
{
if( timediff_2( cTime, m_MatchTime[l] ) == m_Time.at(0) )
{
nearest = m_MatchTime[l];
break;
}
}
}
unsigned int manydaysleft = howmanydaysuntil( nearest.wDayOfWeek , cTime.wDayOfWeek );
unsigned int manyhoursleft = howmanyhoursuntil( nearest.wHour, cTime.wHour );
if( nearest.wHour < cTime.wHour ) //manydaysleft will always be > 0
manydaysleft--;
unsigned int manyminutesleft = howmanyminutesuntil( nearest.wMinute, cTime.wMinute );
if( nearest.wMinute < cTime.wMinute )
manyhoursleft--;
/*cout
<< manydaysleft << endl
<< manyhoursleft << endl
<< manyminutesleft << endl;*/
cout << "CurrentTime\n"
<< "Day:" << cTime.wDayOfWeek
<< "Hour:" << cTime.wHour
<< "Min:" << cTime.wMinute
<< "\nDay:" << nearest.wDayOfWeek
<< "Hour:" << nearest.wHour
<< "Min:" << nearest.wMinute
<< "\nDay:" << manydaysleft
<< "Hour:" << manyhoursleft
<< "Min:" << manyminutesleft;
return 0;
}
答案 4 :(得分：0)
const unsigned n=3; //replace with actual array size
auto packtime = [](const SYSTEMTIME& t)->unsigned
{
return t.wDayOfWeek*24*60 + t.wHour*60 + t.wMinute;
};
auto unpacktime = [](unsigned total)->SYSTEMTIME
{
SYSTEMTIME ret;
ret.wDayOfWeek = total/(60*24);
total %= (60*24);
ret.wHour = total/60;
ret.wMinute = total%60;
return ret;
};
unsigned const wraptime = 7*24*60;
unsigned targettime = packtime(cTime);
unsigned mintimedif = wraptime + 1;
unsigned mindifidx;
unsigned timedif;
for(unsigned i=0; i<n; ++i)
{
timedif = packtime(m_MatchTime[i]);
if(timedif < targettime)
timedif = targettime - timedif;
else
timedif = wraptime - timedif + targettime;
if(timedif < mintimedif)
{
mintimedif = timedif;
mindifidx = i;
}
}
SYSTEMTIME dif = unpacktime(mintimedif);
std::cout<<"Today: Day "<<cTime.wDayOfWeek<<" Hour "<<cTime.wHour<<" Minute "<<cTime.wMinute<<std::endl;
std::cout<<"Nearest day: Day "<<m_MatchTime[mindifidx].wDayOfWeek<<" Hour "<<m_MatchTime[mindifidx].wHour<<" Minute "<<m_MatchTime[mindifidx].wMinute<<std::endl;
std::cout<<"Difference: "<<dif.wDayOfWeek<<" days "<<dif.wHour<<" hours "<<dif.wMinute<<" minutes"<<std::endl;</code>