最大子序列和问题

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#include <iostream> using namespace std;   //暴力求解  T(N)=O(n^3) int  MaxSubSequm1(int list[], int n) {     int MaxSum = 0;     for(int i = 0; i < n; i++)     {         for(int j = i; j < n; j++)         {             int ThisSum = 0;             for(int k = i; k < j; k++)             {                 ThisSum += list[k];             }             if(ThisSum > MaxSum)             {                 MaxSum = ThisSum;             }         }     }     return MaxSum; }   //暴力求解  T(N)=O(n^2) int  MaxSubSequm2(int list[], int n) {     int MaxSum = 0;     for(int i = 0; i < n; i++)     {         int ThisSum = 0;         for(int j = i; j < n; j++)         {             ThisSum += list[j];             if(ThisSum > MaxSum)             {                 MaxSum = ThisSum;             }         }     }     return MaxSum; }   //分而治之 T(N)=O(nlogn) int Maxfun( int A, int B, int C ) {     return (A > B) ? ((A > C) ? A : C) : ((B > C) ? B : C); } int DivideAndConquer(int list[], int left, int right) {     int MaxLeftSum, MaxRightSum; /* 存放左右子问题的解 */     int MaxLeftBorderSum, MaxRightBorderSum; /*存放跨分界线的结果*/        int LeftBorderSum, RightBorderSum;     int center, i;           if(left == right)     {         if(list[left] < 0)         {             return 0;         }         else{             return list[left];         }     }     else{         center = (left + right) / 2;     }           MaxLeftSum = DivideAndConquer(list, left, center);     MaxRightSum = DivideAndConquer(list, center+1, right);           // 求跨分界线的最大子列和     LeftBorderSum = 0; MaxLeftBorderSum = 0;     RightBorderSum = 0; MaxRightBorderSum = 0;     //从中线向左扫描     for(int i = center; i > left; i--)     {         LeftBorderSum += list[i];         if(LeftBorderSum > MaxLeftBorderSum)         {             MaxLeftBorderSum = LeftBorderSum;         }     }     //从中线向右扫描     for(int i = center; i < right; i++)     {         RightBorderSum += list[i];         if(LeftBorderSum > MaxRightBorderSum)         {             MaxRightBorderSum = LeftBorderSum;         }     }           //返回"治"的结果     return Maxfun( MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum ); } int MaxSubSequm3(int list[], int n) {     return DivideAndConquer(list, 0, n-1); }     //在线处理 T(N)=O(n) int  MaxSubSequm4(int list[], int n) {     int ThisSum = 0;     int MaxSum = 0;     for(int i = 0; i < n; i++)     {         ThisSum += list[i];         if(ThisSum > MaxSum)         {             MaxSum = ThisSum;         }         else if(ThisSum < 0)         {             ThisSum = 0;         }     }     return MaxSum; }    void test() {     //int list[6] = { -2, 11, -4, 13, -5, -2 };     int list[1] = {-1};     //int list[9] = {-2,1,-3,4,-1,2,1,-5,4};     int n;     n = sizeof(list) / sizeof(list[0]);     int Maxsum1, Maxsum2, Maxsum3, Maxsum4;     Maxsum1 = MaxSubSequm1(list, n);           Maxsum2 = MaxSubSequm2(list, n);           Maxsum3 = MaxSubSequm3(list, n);           Maxsum4 = MaxSubSequm4(list, n);           cout << Maxsum1 << endl;     cout << Maxsum2 << endl;     cout << Maxsum2 << endl;     cout << Maxsum4 << endl; }   int main() {     test();           system("pause");     return 0; }