舉一反三：

#include <iostream>
#define N 5
#define K 5
using namespace std;

void SearchLatestK();

int arrA[] = {2,5,8,10,11};
int arrB[] = {1,3,5,7,9};
int arrC[K];

int main(int argc, char const *argv[])
{

	SearchLatestK();
	for(int i = 0; i < K; i++)
		cout<<arrC[i]<<" ";
	return 0
 
;
}

void SearchLatestK(){
	int count = 1;
	int i = 0, j = 1, compi = 1, compj = 0;
	arrC[0] = arrA[0] + arrB[0];

	while(count < K){
		if((arrA[i] + arrB[j]) < (arrA[compi] + arrB[compj])){
			arrC[count++] = arrA[i] + arrB[j];//兩陣列首元素相加肯定最小。
			j = ++j == N ? 0,i = compi+1 : j;//這樣的語法也可以，初嘗試。
		}
		else
 
{
			arrC[count++] = arrA[compi] + arrB[compj];
			compj = ++compj == N ? 0,compi = i + 1 : compj;
		}

		if (i == N ){//當其中一個數組走到了盡頭，說明只剩下最後一組和了
			while(count < K) arrC[count++] = arrB[compj];
			return;
		}
		else if( compi == N){
			while(count < K) arrC[count++] = arrB[j];
			return;
		}
	}
}