#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);    	    
}

#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;
}

#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));
}

#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);}

#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_);
}