最大子列和问题的四种算法附世界未解之谜（就是本菜鸡不懂=

#include<iostream> #include<time.h> using namespace std; #define MAXSIZE 100 clock_t start, stop; double duration; int MaxSubseqSum1(int a[], int N) { int ThisSum, MaxSum = 0; int i, j, k; for (i = 0; i < N; i++) { for (j = i; j < N; j++) { ThisSum = 0; for (k = i; k <= j; k++) ThisSum += a[k]; if (ThisSum > MaxSum) MaxSum = ThisSum; } } return MaxSum; } int MaxSubseqSum2(int a[], int N) {// int ThisSum, MaxSum = 0; int i, j; for (i = 0; i < N; i++) { ThisSum = 0; for (j = i; j < N; j++) { ThisSum += a[j]; if (ThisSum > MaxSum) MaxSum = ThisSum; } } return MaxSum; } int Max3(int A, int B, int C) { /* 返回3个整数中的最大值 */ return A > B ? A > C ? A : C : B > C ? B : C; } int DivideAndConquer(int List[], int left, int right) { /* 分治法求List[left]到List[right]的最大子列和 */ 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) { /* 递归的终止条件，子列只有1个数字 */ if (List[left] > 0) return List[left]; else return 0; } /* 下面是"分"的过程 */ center = (left + right) / 2; /* 找到中分点 */ /* 递归求得两边子列的最大和 */ MaxLeftSum = DivideAndConquer(List, left, center); MaxRightSum = DivideAndConquer(List, center + 1, right); /* 下面求跨分界线的最大子列和 */ MaxLeftBorderSum = 0; LeftBorderSum = 0; for (i = center; i >= left; i--) { /* 从中线向左扫描 */ LeftBorderSum += List[i]; if (LeftBorderSum > MaxLeftBorderSum) MaxLeftBorderSum = LeftBorderSum; } /* 左边扫描结束 */ MaxRightBorderSum = 0; RightBorderSum = 0; for (i = center + 1; i <= right; i++) { /* 从中线向右扫描 */ RightBorderSum += List[i]; if (RightBorderSum > MaxRightBorderSum) MaxRightBorderSum = RightBorderSum; } /* 右边扫描结束 */ /* 下面返回"治"的结果 */ return Max3(MaxLeftSum, MaxRightSum, MaxLeftBorderSum + MaxRightBorderSum); } int MaxSubseqSum3(int a[], int N) { return DivideAndConquer(a, 0, N - 1); } int MaxSubseqSum4(int a[], int N) { int ThisSum, MaxSum; int i; ThisSum = MaxSum = 0; for (i = 0; i < N; i++) { ThisSum += a[i]; if (ThisSum > MaxSum) MaxSum = ThisSum; else if (ThisSum < 0) ThisSum = 0; } return MaxSum; } int main() { int n, a[MAXSIZE], i, s = 100000; cout << "input the length of your subsequence:"; cin >> n; cout << "input your subsequence:"; for (i = 0; i < n; i++) cin >> a[i]; start = clock(); for (i = 0; i < s; i++) MaxSubseqSum1(a, n); stop = clock(); duration = ((double)(stop - start)) / CLK_TCK; cout << "first program:" << duration << endl; start = clock(); for (i = 0; i < s; i++) MaxSubseqSum2(a, n); stop = clock(); duration = ((double)(stop - start)) / CLK_TCK; cout << "second program:" << duration << endl; start = clock(); for (i = 0; i < s; i++) MaxSubseqSum3(a, n); stop = clock(); duration = ((double)(stop - start)) / CLK_TCK; cout << "third program:" << duration << endl; start = clock(); for (i = 0; i < s; i++) MaxSubseqSum4(a, n); stop = clock(); duration = ((double)(stop - start)) / CLK_TCK; cout << "forth program:" << duration << endl; }

世界未解之谜，为什么代码运行出来的结果 算法三的分而治之怎么回事？？？

而在PTA提交的结果很对啊

算法2： 算法3： 算法4：为什么？