ZCMU—1133

1133: 第九章：致我们终将逝去的青春

Time Limit: 1 Sec  Memory Limit: 128 MB
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Description

                      

/青 春 是 用 来 追 忆 的

/当 你 怀 揣 着 她 时

/她 一 文 不 值

/只 有 将 她 耗 尽 后

/再 回 过 头 看

/一 切 才 有 了 意 义

/爱 过 我 们 的 人 和 伤 害 过 我 们 的 人

/都 是 我 们 青 春 存 在 的 意 义

         

在ACM中找寻 青春的意义：

Given a specified total t and a list of n integers, find all distinct sums using numbers from the list that add up to t. For example, if t = 4, n = 6, and the list is [4, 3, 2, 2, 1, 1], then there are four different sums that equal 4: 4, 3+1, 2+2, and 2+1+1. (A number can be used within a sum as many times as it appears in the list, and a single number counts as a sum.) Your job is to solve this problem in general.

Input

The input will contain one or more test cases, one per line. Each test case contains t, the total, followed by n, the number of integers in the list, followed by n integers x1, ..., xn. If n = 0 it signals the end of the input; otherwise, t will be a positive integer less than 1000, n will be an integer between 1 and 12 (inclusive), and x1, ..., xn will be positive integers less than 100. All numbers will be separated by exactly one space. The numbers in each list appear in nonincreasing order, and there may be repetitions.

Output

For each test case, first output a line containing 'Sums of', the total, and a colon. Then output each sum, one per line; if there are no sums, output the line 'NONE'. The numbers within each sum must appear in nonincreasing order. A number may be repeated in the sum as many times as it was repeated in the original list. The sums themselves must be sorted in decreasing order based on the numbers appearing in the sum. In other words, the sums must be sorted by their first number; sums with the same first number must be sorted by their second number; sums with the same first two numbers must be sorted by their third number; and so on. Within each test case, all sums must be distinct; the same sum cannot appear twice.

Sample Input

4 6 4 3 2 2 1 1

5 3 2 1 1

400 12 50 50 50 50 50 50 25 25 25 25 25 25

0 0

Sample Output

Sums of 4:

4

3+1

2+2

2+1+1

Sums of 5:

NONE

Sums of 400:

50+50+50+50+50+50+25+25+25+25

50+50+50+50+50+25+25+25+25+25+25

【分析】

数据不大...就是题目长而已...不过题意就算不看题目也看得出来，就是给你一个sum再给你n个数，在这n个数里去找几个数的和等于sum，求出这样所有的等式就可以了，然后给出的n序列是有序的，一个等式中每个数只能用一次。

直接dfs可能还是会超时因为n^n了

加个判重就好了，显然一样的数字在当前这一层循环判断一次就够了，比如第三组那一串50，在某一层搜索中如果用了50这个数字，那么在这一层里就不需要判断后面的50了直接判断25就可以了。因为首先等式不能重复，不判重一定会有重复，除非手动判断等式是否重复，那样的话又白白增加了时间复杂度。其次，不判重也会造成n^n的复杂度导致超时...

【代码】

#include<stdio.h>   int a[2000],ans[2000];  int sum,n,flag;  void dfs(int deep,int len,int m)  {        if(m>sum) return;      if(m==sum)      {          flag=1;          for(int i=0;i<len-1;i++) printf("%d+",ans[i]);          printf("%d\n",ans[len-1]);    }      for(int i=deep;i<n;i++)      {          ans[len]=a[i];          dfs(i+1,len+1,m+a[i]);  while (a[i]==a[i+1]) i++;      }  }  int main()  {      while(~scanf("%d%d",&sum,&n)&&(sum||n))      {          for(int i=0;i<n;i++) scanf("%d",&a[i]);          flag=0;          printf("Sums of %d:\n",sum);          dfs(0,0,0);          if(flag==0) printf("NONE\n");      }      return 0;  }