poj 1564

题意：输出一组数中某几个数之和等于给定的值的所有情况，按降序排列

dfs 加简单的剪枝

题目做的少，递归方程仍然头疼

Memory: 3128K

Time: 297MS

import java.util.Scanner;

public class Main{
    static int n,m,flag;
    static int nums[];
    static int path[];
    
    public static void dfs(int index,int step,int sum){    //index:正在递归的数的下标，step递归的数字个数，sum当前递归的数字和
        if(sum == n){
            flag = 1;
            
            for(int i=0;i<step-1;i++){
                System.out.print(path[i]+"+");
            }
            System.out.println(path[step-1]);
            return;
        }
        
        int pre = -1;                         //记录将要递归的数的前驱，若两个数相同则跳过
        for(int i=index;i<m;i++){
            if(sum+nums[i]<=n&&nums[i]!=pre){
                pre = nums[i];
                path[step] = nums[i];
                dfs(i+1,step+1,sum+nums[i]);
            }
        }
        return;
    }
    
    public static void main(String[] args){
        Scanner in = new Scanner(System.in);
        while(in.hasNext()){
            n = in.nextInt();
            m = in.nextInt();
            if(n==0&&m==0)
                break;
            nums = new int[m];
            path = new int[m];
            for(int i=0;i<m;i++){
                nums[i] = in.nextInt();
            }
            flag = 0;
            System.out.println("Sums of "+n+":");
            dfs(0,0,0);
            if(flag==0){
                System.out.println("NONE");
            }
        }
    }
}