Sum It Up
Time Limit: 1000MS   Memory Limit: 10000K
Total Submissions: 6682   Accepted: 3475

Description

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 x 1 , . . . , x n . 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 x 1 , . . . , x n 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
#include<stdio.h>
#include<string.h>
#define MAX 1100
int n,m,k,ok;
int a[MAX],b[MAX];
void dfs(int pos,int tot,int k)
{
int i,j;
if(tot==n)
{
ok=1;
for(j=0;j<k;j++)
{
if(!j)
printf("%d",b[j]);
else
printf("+%d",b[j]);
}
printf("\n");
return ;
}
for(i=pos;i<m;i++)
{
b[k]=a[i];
dfs(i+1,tot+a[i],k+1);
while(a[i]==a[i+1])//去重
++i;
}
}
int main()
{
int i,j;
while(scanf("%d%d",&n,&m),n|m)
{
for(i=0;i<m;i++)
scanf("%d",&a[i]);
k=0;
ok=0;
printf("Sums of %d:\n",n);
dfs(0,0,0);
if(!ok)
printf("NONE\n");
}
return 0;
}

  

poj 1564 Sum It Up【dfs+去重】的更多相关文章

  1. POJ 1564 Sum It Up (DFS+剪枝)

                                                                                                       ...

  2. poj 1564 Sum It Up(dfs)

    Sum It Up Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 7191   Accepted: 3745 Descrip ...

  3. poj 1564 Sum It Up (DFS+ 去重+排序)

    http://poj.org/problem?id=1564 该题运用DFS但是要注意去重,不能输出重复的答案 两种去重方式代码中有标出 第一种if(a[i]!=a[i-1])意思是如果这个数a[i] ...

  4. poj 1564 Sum It Up

    题目连接 http://poj.org/problem?id=1564 Sum It Up Description Given a specified total t and a list of n ...

  5. poj 1564 Sum It Up | zoj 1711 | hdu 1548 (dfs + 剪枝 or 判重)

    Sum It Up Time Limit : 2000/1000ms (Java/Other)   Memory Limit : 65536/32768K (Java/Other) Total Sub ...

  6. POJ 1564 Sum It Up(DFS)

    Sum It Up Time Limit:1000MS     Memory Limit:32768KB     64bit IO Format:%I64d & %I64u Submit St ...

  7. poj 1564 Sum It Up 搜索

    题意: 给出一个数T,再给出n个数.若n个数中有几个数(可以是一个)的和是T,就输出相加的式子.不过不能输出相同的式子. 分析: 运用的是回溯法.比较特殊的一点就是不能输出相同的式子.这个可以通过ma ...

  8. POJ 1564 经典dfs

    1.POJ 1564 Sum It Up 2.总结: 题意:在n个数里输出所有相加为t的情况. #include<iostream> #include<cstring> #in ...

  9. (深搜)Sum It Up -- poj --1564

    链接: http://poj.org/problem?id=1564 http://acm.hust.edu.cn/vjudge/contest/view.action?cid=88230#probl ...

随机推荐

  1. ios 设置label的高度随着内容的变化而变化

    好吧 步骤1:创建label _GeRenJianJie = [[UILabel alloc]init]; 步骤2:设置label _GeRenJianJie.textColor = RGBAColo ...

  2. GCD的一点理解

    大家都知道GCD 有两种队列:一种是串行队列,一种是并发队列.什么是串行队列?串行队列就是队列中的代码块一个一个按顺序执行,每当上一个代码块执行结束后下一个代码块才会执行.打个比方,如果队列是一些首尾 ...

  3. Mysql 操作手册

    mysql操作手册 版本:5.6.16mysql linux安装基本步骤:#rpm -e --nodeps mysql-lib-5.1.*#rpm -ivh mysql-server#rpm -ivh ...

  4. Qt XML的使用

    Qt中对于XML文件的写入有两种方式,一个是使用QXmlStreamWriter,另一个则为使用Dom.stream流的形式相对来说更加灵活,而且适合处理大文件.Dom方式由于是将内容加载到了内存中进 ...

  5. ubuntu安装配置搜狗拼音输入法

    进入下载目录,在终端执行安装 $sudo dpkg  -i   sogou_pinyin_linux_1.0.0.0033_amd64.deb 安装过程会出现 依赖关系问题 2 修复依赖关系完成搜狗拼 ...

  6. jQuery提供的小方法

    jQuery提供的小方法: 1.选择器 + 事件 + 函数 = 复杂的交互 2.循环处理与选择器匹配的各个元素:each() $("#").each(function(){     ...

  7. angular 控制器之间值得传递

    <div ng-controller="ParentCtrl"> <!--父级--> <div ng-controller="SelfCtr ...

  8. PHPCMS标签大全

    {$head[title]} 页面标题,用法: {$phpcms[sitename]} 网站名称 用法: {$head[keywords]} 要害字 用法: {$head[description]} ...

  9. 2016022608 - redis字符串命令集合

    redis字符串命令: Redis字符串命令用于在Redis管理字符串值.使用Redis字符串命令的语法如下所示: redis 127.0.0.1:6379> COMMAND KEY_NAME ...

  10. java入门(与C++的不同之处)封装篇

    初学java,总是想将它与之前的C++做比较,看了慕客网的java入门视频,一直觉得在面向对象方面,它和C++有太多相同的地方,结果今天学到了两点不同之处,现在将它记录下来: 1.  java的访问修 ...