Swaps the elements in two sequences A and B to minimize the difference between the sum of sequence a and the sum of sequence B.

Http://blog.csdn.net/v_JULY_v/article/details/6126444 32nd

Problem description:

There are two sequences, A and B, all of which are n. The values of the sequence elements are random integers and unordered;
Requirement: by exchanging elements in A and B, the deviation between [Sum of Element A] and [Sum of Element B] is minimized.

CodeImplementation:

# Include <cstdio> # include <cstdlib> # include <cmath> void swapi (int * X, int * Y); int sum (int A [], int N ); int isxina (int x, int A); void process (int A [], int B [], int N); int main () {int A [] = {1, 2, 3, 29, 30}; int B [] = {1,210,232,123 11, 12312}; int n = 5; Process (A, B, n ); for (INT I = 0; I <n; I ++) printf ("% d", a [I]); printf ("\ n "); for (INT I = 0; I <n; I ++) printf ("% d", B [I]); printf ("\ n % d \ n ", ABS (sum (A, n)-sum (B, n); System ("pause"); Return 0;} void swapi (int * X, int * Y) {int temp = * X; * x = * Y; * Y = temp;} int sum (int A [], int N) {int ret = 0; for (INT I = 0; I <n; I ++) RET + = A [I]; return ret;} int isxina (INT X, int) {If (x> 0 & x <A) | (x> A & x <0) return 0; return 1 ;} void process (int A [], int B [], int N) {int sum_a = sum (A, n); int sum_ B = sum (B, n ); int A = sum_a-sum_ B; float minvalue = ABS (A [0]-B [0]-A/2.0); int II = 0, JJ = 0; int flag = 0; if (a = 0) return; For (INT I = 0; I <n; I ++) {for (Int J = 0; j <n; j ++) {int x = A [I]-B [J]; if (x = 0) continue; If (isxina (X, A) = 0) {float temp = (float) ABS (A [I]-B [J]-A/2.0); If (temp <minvalue) {minvalue = temp; II = I; JJ = J; flag = 1 ;}}}if (flag = 1) {swapi (& A [II], & B [JJ]); Process (A, B, n ); // continue solving} // solution thought: // The sum of the current array A and array B is // A = sum (a)-sum (B) /// After the I-th element and the J-th element of B are exchanged, the sum of A and B is // a' = sum () -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 | // assume> 0, // when X is between (0, A), such an exchange can reduce the sum of A and B after the exchange, // The Closer X is to A/2, the better the effect. // If X cannot be found between (0, A), the current A and B are the answer. /// SoAlgorithm Probably as follows: // in A and B, find the I and j that make X Between (0, A) and closest to a/2, and exchange the corresponding I and j elements, // After recalculating A, repeat the previous step until the X Between (0, a) is not found.