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// 1. 題目11 數值的整數次方

// 1.1 沒有考慮負數和0的情況!!!! double Power(double dBase, int iExponent) {double dRes = 1.0;for (int i = 0; i < iExponent; i++){dRes *= dBase;}return dRes; }// 判斷兩個double是否相等 bool Equal(double lhs, double rhs) {if ((lhs - rhs > -0.0000001) && (lhs - rhs < 0.0000001)){return true;}return false; }double PowerWithUnsignedExponent(double dBase, unsigned int uiExponent) {double dResult = 1.0;for (unsigned int i = 1; i <= uiExponent; i++){dResult *= dBase;}return dResult; }// 1.2 全面但不夠高效的算法 double Power_1(double dBase, int iExponent) {double dResult = 0.0;if (Equal(dBase, 0.0) && iExponent < 0){return dResult;}unsigned int absExponent = (unsigned int)iExponent;if (iExponent < 0){absExponent = (unsigned int)(-iExponent);}dResult = PowerWithUnsignedExponent(dBase, absExponent);if (iExponent < 0){dResult = 1.0 / dResult;}return dResult; }// 1.3 全面且高效的解法 // eg 如果輸入iExponent = 32, PowerWithUnsignedExponent中需要循環31次 // 另一種解法: (a)32 --> ((a)16)2 --> (((a)8)2)2 -> ... --> (((((a)2)2)2)2)2 --> 運行5次double NewPowerWithUnsignedExponent(double dBase, unsigned int uiExponent) {if (0 == uiExponent){return 1.0;}else if (1 == uiExponent){return dBase;}double dResult = NewPowerWithUnsignedExponent(dBase, uiExponent >> 1);dResult *= dResult;// 如果是基數if (uiExponent & 0x01){dResult *= dBase;}return dResult; }double Power_2(double dBase, int iExponent) {double dResult = 0.0;if (Equal(dBase, 0.0) && iExponent < 0){return dResult;}unsigned int absExponent = (unsigned int)iExponent;if (iExponent < 0){absExponent = (unsigned int)(-iExponent);}dResult = NewPowerWithUnsignedExponent(dBase, absExponent);if (iExponent < 0){dResult = 1.0 / dResult;}return dResult; }void PowerTestFunc() {cout << "\n\n --------------- PowerTestFunc Start -------------->" << endl;int aiArray[] = {-3, -2, -1, 0, 1, 2, 3, 4, 5, 6};int iLen = sizeof(aiArray) / sizeof(int);double dBase = 5;cout << "方法一: ==========================>>" << endl;for (int i = 0; i < iLen; i++){printf("[%lf] 的 %d 次方: %lf\n", dBase, aiArray[i], Power(dBase, aiArray[i]));}cout << "方法二: ==========================>>" << endl;for (int i = 0; i < iLen; i++){printf("[%lf] 的 %d 次方: %lf\n", dBase, aiArray[i], Power_1(dBase, aiArray[i]));}cout << "方法三: ==========================>>" << endl;for (int i = 0; i < iLen; i++){printf("[%lf] 的 %d 次方: %lf\n", dBase, aiArray[i], Power_2(dBase, aiArray[i]));}cout << "\n\n --------------- PowerTestFunc End -------------->" << endl;}

轉載于:https://www.cnblogs.com/yzdai/p/11258684.html

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