![](/upload/album/3d/51/bb/ee4c35550812GpbajzUk.png)
First:breaktrendline1buy
channel:breakchannelbuy2buy
retestchannel3buy
channelisstrongertrendthantrendline
confirmtrendchange
[
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![](/upload/album/a1/96/25/0f9cb8da5718vvmeo055.png)
12-13:
Timecycle:2monthwindow:
Retrace3000
breakwedgegotochanneltop
pressure
3140break-->moneyout
waitfor300buy
breakchanneltop-->buy1,2,3
Nopressure,riverwidthgood
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![](/upload/album/88/99/a5/25abc48946440xT9Pew5.png)
1.upTrendbuyatthetrendlineorbreakpoint2breakupTrend45out3breakdownTrendlinebreakbuy1,testandupbuy24cutlosswithbuypointoremacrossorbreaksmalltrendline5.bigdowntrendnotbreak,donotbuyBuy2point:breakandbreaktestandupagain1buybreak2buyretestandup3breakbigtrendline[
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![](https://c.mql5.com/2/27/8blp7_l2x8y.png)
Trendline:moststableis4530and70notstableIfgoingupsharplysupportcansetat45trendlinebreakcanreboundorgobackdownbuypoint:trendline[
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![](https://bpcdn.co/images/2016/05/grade1-trendlines-example-thumbnail.png)
Trendline:Thelonger,thevalidAdjustaccordinglybreaktrendline:buyretraceandupagain:buy[
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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
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![](/upload/album/29/f1/23/2daaa9656961VAOuJGxI.png)
Tradetodayaboveopenandbreakdowntrendandaddbreakagaincutlossbelowtrendlineoverall:dowdownopenlowMACDcrossdowndaily1houremacrossdown4hourwaitingcrossdailycrossMarketistop,waitingforsellbutcanbreaktopagainwaitingforabetterbuypointtakeprofitXOPwaitingfor2ndtestgoldwaitingfor12andJanuaryXLEwaitafterJanandFeb[
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![](https://images2015.cnblogs.com/blog/1038557/201706/1038557-20170620112104866-2055715605.jpg)
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[
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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...[
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