poj_3187_Backward Digit Sums

http://poj.org/problem?id=3187

/*总结：头文件#include <algorithm>，
next_permutation(num,num+n)生成数组num的全排列，
*/
#include <iostream>
#include <algorithm>
using namespace std;
int fun(int *num,int n)
{
	int num1[10];//不能直接使用num操作，涉及内存管理，需要用替换数组进行操作
	for(int i = 0;i < n;i++)
		num1[i] = num[i];

	for(int i = 1;i <= n-1;i++)
	{
		int x = 0;
		for(int j = 0;j < n - i;j++)
		{
			num1[x++] = num1[j]+num1[j+1];
		}
	}
	return num1[0];
}
int main(int argc, char *argv[])
{
	int n,finalSum,num[1001];
	while(cin >> n >> finalSum)
	{
		for(int i = 0;i < n;i++)
			num[i] = i+1;
		int sum,cnt = 1;
		do
		{
			sum = fun(num,n);
			if(sum == finalSum)
				break;
		}while(next_permutation(num,num+n));
		for(int i = 0;i < n-1;i++)
			cout << num[i] << " ";
		cout << num[n-1] << "\n";
	}
	return 0;
}