定义一个长度为100的int型数组,数组中前n(
n>=1&&n<=100)个元素已经从小到大有序排列。输入一个整数
,使用二分查找法在该数组中查找和相等的元素。如果在数组中
找到,输出找到的第一个元素的下标;否则,输出“Not Found
#include <stdio.h>
int main()
{
int a[100] = {23, 45, 60, 67, 88}, len = 5;
int value;
int i, type, j;
while (scanf ("%d", &value)){
if (value < 0)break;
for (i = 0; i < len; ++i){
if (value == a[i]){
type = 1;
break;
}
else if (a[i] > value){
type = 2;
break;
}
}
if (i >= len){
a[len++] = value;
}
else {
if (type == 1){
for (j = i; j < len-1; ++j){
a[j] = a[j+1];
}
len--;
}
else if (type == 2){
for (j = len; j > i; --j){
a[j] = a[j-1];
}
len++;
a[i] = value;
}
}
for (i = 0; i < len; ++i){
printf ("%5d", a[i]);
}
printf ("\n");
if (len >= 100){
printf ("More than 100!\n");
break;
}
}
}
这样即可实现二分法找到输入数字所在位置
#include <iostream>
#include <stdio.h>
int dichotomy(int arr[], int len, int value)
{
auto left = 0;
auto right = len - 1;
while (left <= right)
{
auto mid = (left + right) / 2;
if (arr[mid] == value)
return mid;
if (arr[mid] < value)
left = mid + 1;
else
right = mid - 1;
}
return -1;
}
int main()
{
std::cout << "Hello World!\n";
int a[100] = { 23, 45, 60, 67, 88 }, len = 5;
int value;
int i, type, j;
int ret_index = 0;
while (scanf_s("%d", &value)) {
if (value < 0)break;
// 二分法查找输入数字所在位置
ret_index = dichotomy(a, len, value);
if (ret_index != -1)
{
printf("%d\n", ret_index);
}
else
{
printf("Not Found !\n");
}
}
}