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最新推荐文章于 2025-07-28 09:03:04 发布

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最新推荐文章于 2025-07-28 09:03:04 发布

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文章标签：面试 iostream 数据结构 numbers library up

本文链接：https://blog.youkuaiyun.com/lxr215/article/details/2884324

本文提供了一个C++程序示例，该程序用于计算整数数组中连续三个数相加等于24的序列次数。通过遍历数组并累加指定长度内的元素来实现目标，最终返回符合条件的序列数量。

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http://topic.youkuaiyun.com/u/20080905/02/3c98d86c-352e-4ca9-b4ed-e63159805407.html

Using any programming language you like, but without using advanced library functions:

Given an array of integers, count how many times any contiguous sequence of three numbers in the array adds up to the number '24'.

For example, in the following sequence:

8 9 7 4 5 6 11 13 0 4 8

The answer would be: 2, because "8 9 7" adds up to 24 and "11 13 0" adds up to 24.

and, in the following sequence:

12 6 6 8 10 5 4 15 2 7 10

The answer would be: 4.

来不及看数据结构的书了。以上是题目，求高效的算法。
请大家自由发挥~~~~

我的解答：

#include <iostream>
using namespace std;

// length：数组元素个数，sumSeq：和为多少(本例24)， numSeq：序列个数(本例3)
int f(int arry[], int length, int sumSeq, int numSeq);

int main()
{
int a[]={8,9,7, 4, 5, 6, 11, 13, 0, 4, 8};
int b[]={12, 6, 6, 8, 10, 5, 4, 15, 2, 7, 10 };
int lengthA = sizeof(a)/sizeof(a[0]);
int lengthB = sizeof(b)/sizeof(b[0]);
cout < < f(a, lengthA, 24, 3) < <endl;
cout < < f(b, lengthB, 24, 3) < <endl;
}
int f(int arry[], int length, int sumSeq, int numSeq)
{
int count=0;
int sum =0;
int j=0;
int m_numSeq=0;
for(int i=0; i <length-2; ++i)
{
j=i;
sum=0;
m_numSeq = numSeq;
while(m_numSeq!=0)
{
sum += arry[j++];
if(sum > sumSeq)
break;
m_numSeq--;
}
if(sum==sumSeq)
{
count++;
}
}
return count;
}