python求解！！！！ python题目 求解？！！
python\u6c42\u89e3\u8981\u77e5\u90530\u4e3a\u5047\uff0c\u800c\u975e0\u5c31\u4e3a\u771f\u3002
\u800cnumbe%2\u7684\u7ed3\u679c\u53ea\u53ef\u80fd\u662f0\u62161\uff0c\u662f1\u5c31\u4e3a\u771f\uff0c\u800c\u662f1\u5c31\u8bf4\u660enumber\u662f\u5947\u6570\uff0c\u540cnumber%2==1\u662f\u4e00\u4e2a\u610f\u601d\uff0c\u800c\u662f0\u5c31\u4e3a\u5047\u3002
\u5982\u4e0b:
for i in range(1, 18, 2):
xNum = i
hNum = (17 - xNum) / 2
print '{0}{1}{2}'.format('-'*hNum, '*'*xNum, '-'*hNum)
for i in range(15, -1, -2):
xNum = i
hNum = (17 - xNum) / 2
print '{0}{1}{2}'.format('-'*hNum, '*'*xNum, '-'*hNum)
绛旓細鍙互浣跨敤Python鍐呯疆鍑芥暟min鍜宮ax鏉姹傝В鏈灏忓煎拰鏈澶у硷紝鐒跺悗浣跨敤print鍑芥暟鍦ㄥ悓涓琛岃緭鍑轰袱涓暣鏁般備緥濡傦紝濡傛灉瑕佹壘鍑2鍜5涓殑鏈澶у煎拰鏈灏忓硷紝鍙互浣跨敤浠ヤ笅浠ｇ爜锛歛 = 2 b = 5 print("鏈澶у硷細", max(a, b), " 鏈灏忓硷細", min(a, b))杈撳嚭缁撴灉涓猴細鏈澶у硷細 5 鏈灏忓硷細 2 ...
绛旓細Python浠ｇ爜锛歛 = float(input("璇疯緭鍏ヤ笂搴曪細"))b = float(input("璇疯緭鍏ヤ笅搴曪細"))h = float(input("璇疯緭鍏ラ珮锛"))S = (a + b) * h / 2 print("绛夎叞姊舰鐨勯潰绉负锛", S)鍛ㄩ暱锛氳绛夎叞姊舰鐨勪笂搴曚负a锛屼笅搴曚负b锛屾枩杈逛负c锛屽垯鍛ㄩ暱P=a+b+2*c Python浠ｇ爜锛歛 = float(input("璇...
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绛旓細鍦Python缂栫▼璇█涓紝鎴戜滑鍙互浣跨敤while寰幆鏉姹傝В1鍒100涔嬮棿鐨勫伓鏁板拰銆傞鍏堬紝鎴戜滑闇瑕佸垵濮嬪寲涓涓彉閲忔潵琛ㄧず褰撳墠鍋舵暟锛屽苟璁惧畾涓涓疮鍔犲櫒鏉ョ疮璁″伓鏁板拰銆傜劧鍚庯紝鎴戜滑閫氳繃while寰幆锛屾瘡娆″皢鍋舵暟鍔2锛岀劧鍚庣疮鍔犲埌鍋舵暟鍜屼腑锛岀洿鍒板伓鏁拌秴杩100涓烘銆備互涓嬫槸涓涓娇鐢╳hile寰幆姹傝В1鍒100鐨勫伓鏁板拰鐨勭ず渚嬩唬鐮侊細鍒濆鍖栧伓鏁板拰...
绛旓細python缂栧啓涓涓兘姹傝В涓鍏冧簩娆℃柟绋嬬殑灏忕▼搴忔牴鎹笁涓郴鏁版潵姹傝Вx鐨勫...棣栧厛瑕佷簡瑙ｄ竴鍏冧簩娆℃柟宸殑姹傛硶锛岀劧鍚庨愭缂栧啓绋嬪簭銆傛柟绋嬩负锛歛x^2鍔燽x鍔燾绛変簬0鎴戜滑鍏堢紪鍐欎竴涓渶绠鍗曠殑鐗堟湰锛屾垚鍔熺殑璁＄畻闄や簡鏁板笺俻rint(sympy.solve(x**2 + 2*x + 1))spicy.optimize.fsolve姹傛暟鍊艰В锛岄渶瑕佺粰瀹氬垵鍊硷紝蹇呰鏃...
绛旓細1. 绱姞</ 鍦Python涓紝姹傝В1鍒100鐨勫拰鏄熀纭鎿嶄綔锛屾垜浠潵鐪嬬湅涓ょ鏂规硶鐨勫疄鐜帮細鏂规硶涓锛坒or寰幆锛夛細</```python# 瀹氫箟绱姞鍑芥暟def sum_range(start, stop): total = 0 for i in range(start, stop): # range涓嶅寘鎷瑂top total += i return total# 璁＄畻1鍒100鐨勫拰result1 =...
绛旓細1銆佽娉曢敊璇 绋嬪簭鎵ц杩囩▼涓紝python瑙ｉ噴鍣ㄤ細妫娴嬩綘鐨勭▼搴忔槸鍚﹀瓨鍦ㄨ娉曢敊璇紝濡傛灉绋嬪簭鍑洪敊python瑙ｉ噴鍣ㄤ細鎸囧嚭鍑洪敊鐨勪竴琛岋紝骞朵笖鍦ㄦ渶鍏堟壘鍒扮殑锛岄敊璇殑鏂囪亴鏍囪浜嗕竴涓皬灏忕殑绠ご銆2銆侀昏緫閿欒 鍦ㄧ敓娲讳腑0鏄笉鑳戒綔涓鸿闄ゆ暟鐨勶紝绋嬪簭鍐欑殑璇硶鍙兘娌￠棶棰橈紝浣嗘槸閫昏緫涓婁笉涓瀹氭病鏈夐棶棰橈紝杩欏氨鏄竴绉嶉昏緫閿欒銆傚紓甯稿鐞...
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绛旓細鍒ゆ柇鏄惁鏈夊疄鏁拌В if (b ** 2 - 4 * a * c) < 0: # 鏃犲疄鏁拌В print("璇ヤ簩娆″嚱鏁版棤瀹炴暟瑙ｏ紒锛侊紒")else: # 鏈夊疄鏁拌В x1 = round((- b + math.sqrt(b ** 2 - 4 * a * c)) / (2 * a), 2)x2 = round((- b - math.sqrt(b ** 2 - 4 * a * c)) / ...
绛旓細濡備笅:
