两组连续元素的匹配总和

发布于 2025-01-18 00:06:17 字数 504 浏览 5 评论 0原文

给定两组相同长度的整数 P 和 C。如何找到两个集合中相同开始和结束位置的连续元素的所有匹配总和，包括重叠的子集？

首选 C#，但请使用非 Linq。

例子：

Let P and C be the sets of the first ten prime and composite numbers:
P = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 }
C = { 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 }

Match 1: [index starts at 0]
SumP[3..5] = 7 + 11 + 13 = 31
SumC[3..5] = 9 + 10 + 12 = 31

Match 2: [index starts at 0]
SumP[2, 6] = 5 + 7 + 11 + 13 + 17 = 53
SumC[2, 6] = 8 + 9 + 10 + 12 + 14 = 53

我需要判断上面两个和是否是唯一的！

原文

Given two sets of integers P and C of the same length. How can I find ALL matching sums of consecutive elements at the same starting and ending positions in the two sets, including overlapping subsets?

C# is preferred but non-Linq please.

Example:

Let P and C be the sets of the first ten prime and composite numbers:
P = { 2, 3, 5, 7, 11, 13, 17, 19, 23, 29 }
C = { 4, 6, 8, 9, 10, 12, 14, 15, 16, 18 }

Match 1: [index starts at 0]
SumP[3..5] = 7 + 11 + 13 = 31
SumC[3..5] = 9 + 10 + 12 = 31

Match 2: [index starts at 0]
SumP[2, 6] = 5 + 7 + 11 + 13 + 17 = 53
SumC[2, 6] = 8 + 9 + 10 + 12 + 14 = 53

I need to find if the above two sums are the only ones or not!

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赴月观长安 2025-01-25 00:06:18

这是嵌套 for 循环的强力方法，忽略我还不理解的“重叠”部分：

for(int indexA=0; indexA<P.Length; indexA++) {
  for(int indexB=indexA; indexB<P.Length; indexB++) {
    int sumA = 0, sumB = 0;
    for(int i=indexA; i<=indexB; i++) {
      sumA += P[i];
      sumB += C[i];
    }
    if (sumA == sumB) {
      Console.WriteLine("[" + indexA + ", " + indexB + "] : Sum = " + sumA);
    }
  }
}

输出：

[2, 6] : Sum = 53
[3, 5] : Sum = 31

Here's the brute force approach with nested for loops, ignoring the "overlapping" part which I don't understand yet:

for(int indexA=0; indexA<P.Length; indexA++) {
  for(int indexB=indexA; indexB<P.Length; indexB++) {
    int sumA = 0, sumB = 0;
    for(int i=indexA; i<=indexB; i++) {
      sumA += P[i];
      sumB += C[i];
    }
    if (sumA == sumB) {
      Console.WriteLine("[" + indexA + ", " + indexB + "] : Sum = " + sumA);
    }
  }
}

Output: