C语言连续子向量的最大和及时间度量实例

#include <stdio.h>#include <time.h>#include <stdlib.h>#define SCALE 3000int maxnum(int a, int b);int main(int argc, char const *argv[]){ FILE *fp; fp = fopen("maximum.in", "r"); // int x[] = {1,12,-11,10,-65,54,22,-9,21,5,48,5,-8,-2,56,54,-88,-5,2,-8,554,-56,35,-55,555,-65,-545,-23,48,-5,88,-56,16,-8}; int *x = (int *)malloc(sizeof(int)*(SCALE+1)); int xi = SCALE,a = 0,num_in = 0; while(xi--){  fscanf(fp, "%d", &x[a++]); } clock_t start, end; // ***Algorithm-1 cube*** start = clock(); int max = 0; int length = SCALE; int i,j,k; for (i = 0; i < length; ++i) {  for (j = i; j < length; ++j)  {   int sum = 0;   for (k = i; k <= j; ++k)   {    sum += x[k];   }   max = maxnum(max, sum);  } }  // long num = 10000000L; // while(num--); end = clock(); double times = (double)(end - start)/CLOCKS_PER_SEC; double dend = (double)end; printf("/n***Algorithm-1 cube***/n"); printf("end: %f/n", dend); printf("Time consuming: %f/n", times); printf("%d/n", max); // ***Algorithm-2 square*** start = clock(); max = 0; for (i = 0; i < length; ++i) {  int sum = 0;  for (j = i; j < length; ++j)  {   sum += x[j];   max = maxnum(max, sum);  } } end = clock(); times = (double)(end - start)/CLOCKS_PER_SEC; dend = (double)end; printf("/n***Algorithm-2 square***/n"); printf("end: %f/n", dend); printf("Time consuming: %f/n", times); printf("%d/n", max); // ***Algorithm-3 linear*** start = clock(); max = 0; int max_end_here = 0; for (i = 0; i < length; ++i) {  max_end_here = maxnum(max_end_here + x[i], 0);  max = maxnum(max, max_end_here); } end = clock(); times = (double)(end - start)/CLOCKS_PER_SEC; dend = (double)end; printf("/n***Algorithm-3 linear***/n"); printf("end: %f/n", dend); printf("Time consuming: %f/n", times); printf("%d/n", max); free(x); x = NULL; return 0;}int maxnum(int a, int b){ return a > b ? a : b;}