First:breaktrendline1buy channel:breakchannelbuy2buy retestchannel3buy channelisstrongertrendthantrendline confirmtrendchange [阅读全文]
大浪淘沙
以平常心对股市沉浮, 不悔不怕
博文
(2019-12-13 10:13:45)
First:breaktrendline1buy channel:breakchannelbuy2buy retestchannel3buy channelisstrongertrendthantrendline confirmtrendchange [阅读全文]
(2019-12-13 08:49:38)
12-13: Timecycle:2monthwindow: Retrace3000 breakwedgegotochanneltop pressure 3140break-->moneyout waitfor300buy breakchanneltop-->buy1,2,3 Nopressure,riverwidthgood [阅读全文]
(2019-12-13 00:27:09)
1.upTrendbuyatthetrendlineorbreakpoint2breakupTrend45out3breakdownTrendlinebreakbuy1,testandupbuy24cutlosswithbuypointoremacrossorbreaksmalltrendline5.bigdowntrendnotbreak,donotbuyBuy2point:breakandbreaktestandupagain1buybreak2buyretestandup3breakbigtrendline[阅读全文]
(2019-12-12 23:48:24)
Trendline:moststableis4530and70notstableIfgoingupsharplysupportcansetat45trendlinebreakcanreboundorgobackdownbuypoint:trendline[阅读全文]
(2019-12-12 23:22:48)
(2019-12-12 22:48:58)
Basic:
Global:tradewar
Area:war
PE
Finance:interest,tax
Supply/demand:oil,naturalgas:weather,gold:chinesenewyear
Policy
Technique:
Dow:
Trendline,shape,position
Jianen28,time(2months)
Position,box
Indicator:
MACDemabollingerband
[阅读全文]
(2019-12-12 09:48:45)
(2019-12-11 10:00:06)
Tradetodayaboveopenandbreakdowntrendandaddbreakagaincutlossbelowtrendlineoverall:dowdownopenlowMACDcrossdowndaily1houremacrossdown4hourwaitingcrossdailycrossMarketistop,waitingforsellbutcanbreaktopagainwaitingforabetterbuypointtakeprofitXOPwaitingfor2ndtestgoldwaitingfor12andJanuaryXLEwaitafterJanandFeb[阅读全文]
(2019-12-05 10:25:04)
题意及分析：根据题目要求，数组里的每个元素表示从该位置可以跳出的最远距离，要求问从第一个元素（index=0）开始，能否达到数组的最后一个元素，这里认为最后一个元素为终点。这里是到达，说明超过也行，看下图能更好的理解题意。所以这里可以使用贪心算法，计算出某个点时能够跳出的最大距离（当前的最大值和（当前点+能跳出的最大距离）的较大的值），如果能[阅读全文]
(2019-12-05 10:11:09)
Knowledgebase:1.matrix:doubleloop2dpsimilartoonedimensionfirstinitializethevaluethenthedp[i][j]=dp[i-1][j]+dp[i][j-1];publicintuniquePaths(intm,intn){
int[][]dp=newint[m][n];
for(inti=0;i<m;i++){
dp[i][0]=1;
}
for(intj=0;j<n;j++){
dp[0][j]=1;
}
for(inti=1;i<m;i++){
for(intj=1;j<n;j++){
dp[i][j]=dp[i][j-1]+dp[i-1][j];
}
}
returndp...[阅读全文]
int[][]dp=newint[m][n];
for(inti=0;i<m;i++){
dp[i][0]=1;
}
for(intj=0;j<n;j++){
dp[0][j]=1;
}
for(inti=1;i<m;i++){
for(intj=1;j<n;j++){
dp[i][j]=dp[i][j-1]+dp[i-1][j];
}
}
returndp...[阅读全文]
