안녕하세요 예스스탁입니다. input : n(3); var : YO(0),wm(0),cnt(0),HH(0),LL(0); Array : YH[20](0),YL[20](0); if bdate > bdate[1]+365 Then { YO = O; YH[0] = H; YL[0] = L; wm=(YH[1]+YL[1])/2; For cnt = 1 to 19 { YH[cnt] = YH[cnt-1][1]; YL[cnt] = YL[cnt-1][1]; } } if YH[0] > 0 and H > YH[0] Then YH[0] = H; if YL[0] > 0 and L < YL[0] Then YL[0] = L; if YH[n-1] > 0 and YL[n-1] > 0 Then { HH = YH[0]; LL = LL[0]; For cnt = 1 to n-1 { if YH[cnt] > HH Then HH = YH[cnt]; if YL[cnt] < LL Then LL = YL[cnt]; } var15 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(11.8/100)); var10 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(23.6/100)); var20 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(38.2/100)); var30 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(50/100)); var40 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(61.8/100)); var50 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(76.4/100)); var65 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(88.2/100)); } 즐거운 하루되세요 > 뮬리 님이 쓴 글입니다. > 제목 : 부탁드립니다. > 도움에 감사 드립니다. 1년 기간의 피보나치선을 예를들어 월봉에서 3년또는 5년기간의 피보나치선이 가능 한지요? var : hh30(0),ll30(0),mo(0),wm(0); if bdate > bdate[1]+365 Then{ MO = O; hh30 = H; ll30 = L; wm=(hh30[1]+ll30[1])/2; } if H > hh30 Then hh30 = H; if L < ll30 Then ll30 = L; var15 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(11.8/100)); var10 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(23.6/100)); var20 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(38.2/100)); var30 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(50/100)); var40 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(61.8/100)); var50 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(76.4/100)); var65 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(88.2/100));

답변수식 감사 드립니다. 그림과 같이 실행결과 정상적이지 않습니다. 수정 부탁 드립니다. input : n(3); var : YO(0),wm(0),cnt(0),HH(0),LL(0); Array : YH[20](0),YL[20](0); if bdate > bdate[1]+365 Then { YO = O; YH[0] = H; YL[0] = L; wm=(YH[1]+YL[1])/2; For cnt = 1 to 19 { YH[cnt] = YH[cnt-1][1]; YL[cnt] = YL[cnt-1][1]; } } if YH[0] > 0 and H > YH[0] Then YH[0] = H; if YL[0] > 0 and L < YL[0] Then YL[0] = L; if YH[n-1] > 0 and YL[n-1] > 0 Then { HH = YH[0]; LL = LL[0]; For cnt = 1 to n-1 { if YH[cnt] > HH Then HH = YH[cnt]; if YL[cnt] < LL Then LL = YL[cnt]; } var15 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(11.8/100)); var10 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(23.6/100)); var20 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(38.2/100)); var30 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(50/100)); var40 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(61.8/100)); var50 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(76.4/100)); var65 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(88.2/100)); } plot91(hh,"H",yellow,def,2); plot92(var15,"11.8",yellow); plot93(var10,"23.6",yellow); plot94(var20,"38.2",yellow); plot95(var30,"50.0",magenta,def,2); plot96(var40,"61.8",green); plot97(var50,"76.4",green); plot98(var65,"88.2",green); plot99(ll,"L",green,def,2); > 예스스탁 님이 쓴 글입니다. > 제목 : Re : 부탁드립니다. > 안녕하세요 예스스탁입니다. input : n(3); var : YO(0),wm(0),cnt(0),HH(0),LL(0); Array : YH[20](0),YL[20](0); if bdate > bdate[1]+365 Then { YO = O; YH[0] = H; YL[0] = L; wm=(YH[1]+YL[1])/2; For cnt = 1 to 19 { YH[cnt] = YH[cnt-1][1]; YL[cnt] = YL[cnt-1][1]; } } if YH[0] > 0 and H > YH[0] Then YH[0] = H; if YL[0] > 0 and L < YL[0] Then YL[0] = L; if YH[n-1] > 0 and YL[n-1] > 0 Then { HH = YH[0]; LL = LL[0]; For cnt = 1 to n-1 { if YH[cnt] > HH Then HH = YH[cnt]; if YL[cnt] < LL Then LL = YL[cnt]; } var15 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(11.8/100)); var10 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(23.6/100)); var20 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(38.2/100)); var30 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(50/100)); var40 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(61.8/100)); var50 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(76.4/100)); var65 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(88.2/100)); } 즐거운 하루되세요 > 뮬리 님이 쓴 글입니다. > 제목 : 부탁드립니다. > 도움에 감사 드립니다. 1년 기간의 피보나치선을 예를들어 월봉에서 3년또는 5년기간의 피보나치선이 가능 한지요? var : hh30(0),ll30(0),mo(0),wm(0); if bdate > bdate[1]+365 Then{ MO = O; hh30 = H; ll30 = L; wm=(hh30[1]+ll30[1])/2; } if H > hh30 Then hh30 = H; if L < ll30 Then ll30 = L; var15 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(11.8/100)); var10 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(23.6/100)); var20 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(38.2/100)); var30 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(50/100)); var40 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(61.8/100)); var50 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(76.4/100)); var65 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(88.2/100));

안녕하세요 예스스탁입니다. 수정한 식입니다. input : n(3); var : YO(0),wm(0),cnt(0),HH(0),LL(0); Array : YH[20](0),YL[20](0); if bdate > bdate[1]+365 Then { YO = O; YH[0] = H; YL[0] = L; wm=(YH[1]+YL[1])/2; For cnt = 1 to 19 { YH[cnt] = YH[cnt-1][1]; YL[cnt] = YL[cnt-1][1]; } } if YH[0] > 0 and H > YH[0] Then YH[0] = H; if YL[0] > 0 and L < YL[0] Then YL[0] = L; if YH[n-1] > 0 and YL[n-1] > 0 Then { HH = YH[0]; LL = YL[0]; For cnt = 1 to n-1 { if YH[cnt] > HH Then HH = YH[cnt]; if YL[cnt] < LL Then LL = YL[cnt]; } var15 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(11.8/100)); var10 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(23.6/100)); var20 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(38.2/100)); var30 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(50/100)); var40 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(61.8/100)); var50 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(76.4/100)); var65 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(88.2/100)); plot91(hh,"H",yellow,def,2); plot92(var15,"11.8",yellow); plot93(var10,"23.6",yellow); plot94(var20,"38.2",yellow); plot95(var30,"50.0",magenta,def,2); plot96(var40,"61.8",green); plot97(var50,"76.4",green); plot98(var65,"88.2",green); plot99(ll,"L",green,def,2); } 즐거운 하루되세요 > 뮬리 님이 쓴 글입니다. > 제목 : Re : Re : 부탁드립니다. > 답변수식 감사 드립니다. 그림과 같이 실행결과 정상적이지 않습니다. 수정 부탁 드립니다. input : n(3); var : YO(0),wm(0),cnt(0),HH(0),LL(0); Array : YH[20](0),YL[20](0); if bdate > bdate[1]+365 Then { YO = O; YH[0] = H; YL[0] = L; wm=(YH[1]+YL[1])/2; For cnt = 1 to 19 { YH[cnt] = YH[cnt-1][1]; YL[cnt] = YL[cnt-1][1]; } } if YH[0] > 0 and H > YH[0] Then YH[0] = H; if YL[0] > 0 and L < YL[0] Then YL[0] = L; if YH[n-1] > 0 and YL[n-1] > 0 Then { HH = YH[0]; LL = LL[0]; For cnt = 1 to n-1 { if YH[cnt] > HH Then HH = YH[cnt]; if YL[cnt] < LL Then LL = YL[cnt]; } var15 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(11.8/100)); var10 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(23.6/100)); var20 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(38.2/100)); var30 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(50/100)); var40 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(61.8/100)); var50 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(76.4/100)); var65 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(88.2/100)); } plot91(hh,"H",yellow,def,2); plot92(var15,"11.8",yellow); plot93(var10,"23.6",yellow); plot94(var20,"38.2",yellow); plot95(var30,"50.0",magenta,def,2); plot96(var40,"61.8",green); plot97(var50,"76.4",green); plot98(var65,"88.2",green); plot99(ll,"L",green,def,2); > 예스스탁 님이 쓴 글입니다. > 제목 : Re : 부탁드립니다. > 안녕하세요 예스스탁입니다. input : n(3); var : YO(0),wm(0),cnt(0),HH(0),LL(0); Array : YH[20](0),YL[20](0); if bdate > bdate[1]+365 Then { YO = O; YH[0] = H; YL[0] = L; wm=(YH[1]+YL[1])/2; For cnt = 1 to 19 { YH[cnt] = YH[cnt-1][1]; YL[cnt] = YL[cnt-1][1]; } } if YH[0] > 0 and H > YH[0] Then YH[0] = H; if YL[0] > 0 and L < YL[0] Then YL[0] = L; if YH[n-1] > 0 and YL[n-1] > 0 Then { HH = YH[0]; LL = LL[0]; For cnt = 1 to n-1 { if YH[cnt] > HH Then HH = YH[cnt]; if YL[cnt] < LL Then LL = YL[cnt]; } var15 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(11.8/100)); var10 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(23.6/100)); var20 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(38.2/100)); var30 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(50/100)); var40 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(61.8/100)); var50 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(76.4/100)); var65 = 10^(LOG10(HH)-(LOG10(HH)-LOG10(LL))*(88.2/100)); } 즐거운 하루되세요 > 뮬리 님이 쓴 글입니다. > 제목 : 부탁드립니다. > 도움에 감사 드립니다. 1년 기간의 피보나치선을 예를들어 월봉에서 3년또는 5년기간의 피보나치선이 가능 한지요? var : hh30(0),ll30(0),mo(0),wm(0); if bdate > bdate[1]+365 Then{ MO = O; hh30 = H; ll30 = L; wm=(hh30[1]+ll30[1])/2; } if H > hh30 Then hh30 = H; if L < ll30 Then ll30 = L; var15 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(11.8/100)); var10 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(23.6/100)); var20 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(38.2/100)); var30 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(50/100)); var40 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(61.8/100)); var50 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(76.4/100)); var65 = 10^(LOG10(hh30)-(LOG10(hh30)-LOG10(ll30))*(88.2/100));