There are two arrays A and B whose sizes are N. By exchanging elements in A and B, sum (a)-sum (B) is minimized.

There are two arrays A and B, all of which are N in size. The values of the array elements are unordered due to arbitrary integer numbers;
Requirement: by exchanging elements in A and B, the difference between the sum of Elements A and B is minimized.

The sum of the current array A and array B is
A = sum (a)-sum (B)

After the I-th element of A is exchanged with the J-th element of B, the sum of A and B is
A' = sum (a)-A [I] + B [J]-(sum (B)-B [J] + A [I])
= Sum (a)-sum (B)-2 (A [I]-B [J])
= A-2 (A [I]-B [J])
Set x = A [I]-B [J]

| A |-| a' | = | A |-| A-2x |

Suppose a> 0,

When X is between (0, A), this exchange can reduce the sum of A and B after the exchange. The closer X is to A/2, the better the effect,

If no X is found between (0, A), the current A and B are the answer.

The algorithm is roughly as follows:

Find the I and j that make X Between (0, A) and closest to A/2 in A and B, exchange the corresponding I and j elements, and recalculate, repeat the preceding steps until the X Between (0, a) is not found.

Int sum1, sum2, A; // respectively represents the sum of a [] and B [], and the difference of 2: int temp; bool dayu0; // whether the difference between the sum and is greater than 0 int pos1 and pos2; // subscripts of a [I] and B [J] awaiting exchange I j float Minn; // A [I]-B [J] value bool have1 closest to A/2; // whether the while (1) {sum1 = 0; sum2 = 0; for (INT I = 0; I <n; ++ I) // calculate the sum of the two arrays {sum1 + = AR1 [I]; sum2 + = AR2 [I];} A = sum1-sum2; // The difference between dayu0 = a> 0? True: false; // whether the difference is greater than 0 or less than 0 have1 = false; // whether the solution can be found for (INT I = 0; I <n; ++ I) // find the [I]-B [J] {for (Int J = 0; j <n; ++ J) closest to A/2) {temp = AR1 [I]-AR2 [J]; If (dayu0 & temp> 0 & temp <A) | (! Dayu0 & temp <0 & temp> A) // If a [I]-B [J] is between (0,) {If (have1 & ABS (temp-A/2.0) <Minn) // if it is closer to a/2 than a [I]-B [J], update {Minn = ABS (temp-A/2.0); pos1 = I; pos2 = J ;}else {have1 = true; Minn = ABS (temp-A/2.0); pos1 = I; pos2 = J ;}}} if (! Have1) // if no matching a [I]-B [J] is found, {break;} swap (AR1 [pos1], AR2 [pos2]) ends. // exchange elements in a B}

　　

There are two arrays A and B whose sizes are N. By exchanging elements in A and B, sum (a)-sum (B) is minimized.