本文实例分析了C语言连续子向量的最大和及时间度量，分享给大家供大家参考之用。具体方法如下：
#include <stdio.h> #include <time.h> #include <stdlib.h> #define SCALE 3000 int 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: %fn", dend); printf("Time consuming: %fn", times); printf("%dn", 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: %fn", dend); printf("Time consuming: %fn", times); printf("%dn", 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: %fn", dend); printf("Time consuming: %fn", times); printf("%dn", max); free(x); x = NULL; return 0; } int maxnum(int a, int b) { return a > b ? a : b; }
感兴趣的朋友可以测试运行一下本文实例以加深理解。希望本文所述对大家C程序设计的学习有所帮助。
【C语言连续子向量的最大和及时间度量实例】相关文章：