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;}