数据结构(C语言版)第2版习题答案—严蔚敏

2015.3II1..................................................................................................................12..............................................................................................................53.........................................................................................................144.........................................................................................275.....................................................................................................346......................................................................................................................447................................................................................................................558................................................................................................................66II1..................................................................................................................12..............................................................................................................53.........................................................................................................144.........................................................................................275.....................................................................................................346......................................................................................................................447................................................................................................................558................................................................................................................66222521110025A110B108C100D120B5100+2*4=1082nO(1)Ai1ini2inBi1inCi1inDnAO(n2)O(n2)O(nlog2)iiO(1)3127A8B63.5C63D7Bn/24ABCDA5ABCDD6ACB67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(D+N)18nAnB2n-1C2nDn-1An9ni1in+1An-iBn-i+1Cn-i-1DIB(10)L=(a1a2,an)ABCDD(11)AO(1)BO(n)CO(n2)DO(nlog2n)CO(n)O(n2)(12)ABCDD(13)sp67A1B1C1DCDND/(