【0】README

1)本文旨在review 算法分析的几个算法问题 附带源码；

【1】最大子序列和问题的解（两种解法——分治法+联机算法（推荐））

【1.1】分治法

1）intro：其思想是把问题分成两个大致相等的子问题，然后递归地对它们进行求解，这是“分”部分；“治”阶段 将两个子问题的 解合并到一起并可能再做些少量的附加工作，最后得到整个问题的解；
2）算法idea 描述：在我们的例子中（求最大子序列和的问题），最大子序列和可能出现在三个地方——或者整个出现在输入数据的左半部、或者整个出现在

右半部、或者跨越输入数据的中部从而占据左右两半部分；

3）如何求其解呢？  前两种情况，可以递归求解，第3种情况的最大和可以通过求出前半部分的最大和（包含前半部分的最后一个元素）以及 后半部分的最大 和（包含后半部分的第一个元素）而得到。然后将这两个和加在一起；

4）算法流程演示和源码如下：

#include<stdio.h>#define ElementType intint max3(int i, int j, int k)
{if(i < j){i = j;}if(i < k){i = k;}return i;
}// compute the maximum sum of sebsequence.
// 3 in mxSubSum3 means 3 possibilities.
int maxSubSum3(int A[], int left, int right)
{int onlyLeftSum, onlyRightSum; // just only left or right sum without center.int centerLeftSum, centerRightSum; // the 3rd is passing center.int tempSum;int center, i;if(left==right){if(A[left] > 0){return A[left];}else{return 0;}}center = (left+right)/2;  onlyLeftSum = maxSubSum3(A, left, center);onlyRightSum = maxSubSum3(A, center + 1, right);centerLeftSum = 0;    tempSum = 0;    for(i=center; i>=left; i--) {   tempSum += A[i];if(tempSum > centerLeftSum){centerLeftSum = tempSum;}}centerRightSum = 0;tempSum = 0;for(i=center+1; i<=right; i++) { tempSum += A[i];if(tempSum > centerRightSum){centerRightSum = tempSum;}}printf("onlyLeftSum=%d, onlyRightSum=%d, centerLeftSum + centerRightSum=%d \n", onlyLeftSum, onlyRightSum, centerLeftSum + centerRightSum);return max3(onlyLeftSum, onlyRightSum, centerLeftSum + centerRightSum);
} int main()
{int N = 8, maxSum = 0;            ElementType A[] = {4, -3, 5, -2, -1, 2, 6, -2};        maxSum = maxSubSum3(A, 0, N-1);        printf("the maximum sum of array {4, -3, 5, -2, -1, 2, 6, -2} is: %d \n", maxSum);
}

【1.2】联机算法

1）intro：在任意时刻，算法对要操作的数据只需要读入一次，一旦被读入处理，它就不再需要被记忆了。而在处理过程中，算法对已读入的数据给出了正确的答案，具有这种特性的算法叫联机算法；

2）利用 联机算法求最大子序列和问题的解，源码如下（比 分治法效率高）

#include<stdio.h>#define ElementType int// 联机算法计算最大子序列和问题.
int maxSubSequenceSum(int A[], int N)
{int thisSum=0, maxSum=0;int j;    for(j = 0; j < N; j++) {thisSum += A[j];if(thisSum > maxSum){maxSum = thisSum;}else if(thisSum < 0){thisSum = 0;}}return maxSum;
}int main()
{int N = 8, maxSum = 0;            ElementType A[] = {4, -3, 5, -2, -1, 2, 6, -2};                    maxSum = maxSubSequenceSum(A, N);printf("the maximum sum of array  {4, -3, 5, -2, -1, 2, 6, -2} is %d \n", maxSum);    return 0;
}

【2】运行时间中的对数

【2.1】荔枝1——二分查找 

#include<stdio.h>#define ElementType int
#define NOTFOUND -1/* 二分查找，A[]升序排列 */
int binarySearch(ElementType A[], ElementType x, int N)
{int low, mid, high;low = 0;high = N - 1;while (low <= high){mid = (low + high) / 2;if(A[mid] < x){low = mid + 1;}else if(A[mid] > x){high = mid - 1; }else {return mid;}}return NOTFOUND;
} main()
{int A[]={4, 5, 67, 67, 109, 876};int N=6, x=109;        printf("\narray[] = {4, 5, 67, 67, 109, 876}");printf("\nthe position whose value equals to %d is %4d\n", x, binarySearch(A, x, N));
}

【2.1】荔枝2——计算最大公因数（欧几里得 算法）

#include<stdio.h>#define ElementType int
#define NOTFOUND -1/* 求两个数的最大公因数（greatest common divisor） */
int gcd(int M, int N)
{int rem;printf("remainder sequence: ");while (N > 0){rem = M % N;M = N;N = rem;printf("%4d", rem);}return M;
}main()
{int N = 1989, M = 1590;    int gcd_value;printf("\t\t\t ========== test for greatest common divisor ==========\n\n");    gcd_value = gcd(M, N);printf("\nthe greatest common divisor between %4d and %4d is %4d\n", M, N, gcd_value);}

【2.1】荔枝3——高效率的幂运算

1）减少乘法次数：如 X^62 只用到了9次乘法：

X^3=(X^2)*X , X^7=(X^3)^2*X , X^15=(X^7)^2*X , X^31=(X^15)^2*X , X^62=(X^31)^2

2 + 2 + 2 + 2 + 1 == 9次； 

#include<stdio.h>int isEven(int N)
{return N % 2 == 0 ? 1 : 0;
}int pow(int x, int N)
{if(N == 0){return 1;}else if(N == 1){return x;}else if(isEven(N))    /* if(x) == if(N!=0) */{return pow(x * x, N / 2);}else{return pow(x * x, N / 2) * x;}
}void main()
{int x = 2, N = 7;long pow_ = pow(x,N);printf("\t\t\t ========== test for computing pow  ==========\n");printf("\t\t\t result of %d^%d  is %-6d\n", x, N, pow_);
}