练习1:求1!+2!+...+10!
一般算法:双层循环,外层1~10,内层计算每个数的阶乘,在外层把阶乘相加。
int main() { int i = 0; int j = 0; int jc = 1; int sum = 0; for (i = 1; i <= 10; i++) { jc = 1;// for (j = 1; j <= i; j++) { jc = jc * j; } sum = sum + jc; } printf("%d %d\n", jc,sum); return 0; }
算法优化:1~10递增,其阶乘比上一个多乘一个,在这过程中相加,一层循环即可。
int main() { int i = 0; int jc = 1; int sum = 0; for (i = 1; i <= 10; i++) { jc = jc * i; sum = sum + jc; } printf("%d %d\n", jc, sum); return 0; }
练习2:有序数组查找, 报数,在数组中查找,输出其下标
一般算法:遍历数组,if条件判断输出。
int main() { int input = 0; scanf("%d", &input); int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 }; int len = sizeof(arr) / sizeof(arr[0]); //求出数组长度 printf("%d\n", sizeof(arr)); printf("len=%d\n", len); int i = 0; for (i = 0; i < len; i++) { if (input == arr[i]) { printf("下标为:%d\n", i); break; } } if (i == len) { printf("i=%d\n",i); printf("out of this area"); } return 0; }
未能利用 “有序” 这个条件
算法优化:一层循环用 二分法 折半查找,每次查找少一半数据,效率高。
int main() { int input = 0; scanf("%d", &input); int arr[] = { 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 }; int len = sizeof(arr) / sizeof(arr[0]); //求出数组长度 printf("%d\n", sizeof(arr)); printf("len=%d\n", len); int left = 0; int right = len - 1; int mid = 0; while (left <= right) { mid = (left + right) / 2; if (arr[mid] < input) { left = mid + 1; } if (arr[mid] > input) { right = mid - 1; } if (input == arr[mid]) { printf("找到了,下标为:%d\n", mid); break; } } if (left > right) printf("out of this area"); return 0; }
标签:arr,int,第十篇,笔记,算法,len,printf,input,sizeof From: https://www.cnblogs.com/xiaowanglong/p/17892561.html