http://acm.hdu.edu.cn/showproblem.php?pid=1258

Sum It Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others) Total Submission(s): 4564    Accepted Submission(s): 2349

Problem 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 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 distince; the same sum connot 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<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int t,n,a[],vis[],b[];
int flag;
void dfs(int x,int cur,int sum)
{
int i;
if(sum==t)
{
flag=;
printf("%d",b[]);
for(i=;i<cur;i++)
printf("+%d",b[i]);
printf("\n");
return ;
}
for(i=x;i<n;i++)
{
if(vis[i]==&&sum<t)
{
vis[i]=;
b[cur]=a[i];
dfs(i+,cur+,sum+a[i]);
vis[i]=;
while(i+<n&& a[i]==a[i+]) //去除相同的分解式
i++;
}
}
}
int main()
{
int i;
while(~scanf("%d%d",&t,&n))
{
if(t==&&n==)
break;
flag=;
for(i=;i<n;i++)
scanf("%d",&a[i]);
printf("Sums of %d:\n",t);
dfs(,,);
//printf("flag=%d\n",flag);
if(flag==)
printf("NONE\n");
}
return ; }

Sum It Up的更多相关文章

  1. LeetCode - Two Sum

    Two Sum 題目連結 官網題目說明: 解法: 從給定的一組值內找出第一組兩數相加剛好等於給定的目標值,暴力解很簡單(只會這樣= =),兩個迴圈,只要找到相加的值就跳出. /// <summa ...

  2. Leetcode 笔记 113 - Path Sum II

    题目链接:Path Sum II | LeetCode OJ Given a binary tree and a sum, find all root-to-leaf paths where each ...

  3. Leetcode 笔记 112 - Path Sum

    题目链接:Path Sum | LeetCode OJ Given a binary tree and a sum, determine if the tree has a root-to-leaf ...

  4. POJ 2739. Sum of Consecutive Prime Numbers

    Sum of Consecutive Prime Numbers Time Limit: 1000MS   Memory Limit: 65536K Total Submissions: 20050 ...

  5. BZOJ 3944 Sum

    题目链接:Sum 嗯--不要在意--我发这篇博客只是为了保存一下杜教筛的板子的-- 你说你不会杜教筛?有一篇博客写的很好,看完应该就会了-- 这道题就是杜教筛板子题,也没什么好讲的-- 下面贴代码(不 ...

  6. [LeetCode] Path Sum III 二叉树的路径和之三

    You are given a binary tree in which each node contains an integer value. Find the number of paths t ...

  7. [LeetCode] Partition Equal Subset Sum 相同子集和分割

    Given a non-empty array containing only positive integers, find if the array can be partitioned into ...

  8. [LeetCode] Split Array Largest Sum 分割数组的最大值

    Given an array which consists of non-negative integers and an integer m, you can split the array int ...

  9. [LeetCode] Sum of Left Leaves 左子叶之和

    Find the sum of all left leaves in a given binary tree. Example: 3 / \ 9 20 / \ 15 7 There are two l ...

  10. [LeetCode] Combination Sum IV 组合之和之四

    Given an integer array with all positive numbers and no duplicates, find the number of possible comb ...

随机推荐

  1. 富客户端 wpf, Winform 多线程更新UI控件

    前言 在富客户端的app中,如果在主线程中运行一些长时间的任务,那么应用程序的UI就不能正常相应.因为主线程要负责消息循环,相应鼠标等事件还有展现UI. 因此我们可以开启一个线程来格外处理需要长时间的 ...

  2. 【电视桌面CSWUI】电视桌面(launcher)截图欣赏

    网络播放器是最重要的电视桌面.cswui,我们公司做了一个非常大的人力,物力搞一个电视柜.后来一一介绍,简言之发送屏幕截图.给大家看. watermark/2/text/aHR0cDovL2Jsb2c ...

  3. Oracle中sys和system用户的区别

    1.数据库的启动需要以SYSDBA/SYSOPER身份登录. 2.如果在同一主机上使用IPC连接到数据库使用操作系统授权,登录任何一个用户都可以拥有as sysdba和as sysoper. 3.sy ...

  4. Myeclipse 配置Tomcat 出现 “Value must be an existing directory”错误

    今天上午配了一下本机上的Myeclipse的tomcat,因为我本机上有两个版本的myeclipse,一个是用来公司开发的,一个是自己玩的,本机上装了两个版本jdk和两个版本的tomcat.配置自己玩 ...

  5. android 开发从入门到精通

    Android-Tips This is an awesome list of tips for android. If you are a beginner, this list will be t ...

  6. Win10系列:C#应用控件进阶8

    LineGeometry LineGeometry控件通过指定直线的起点和终点来定义线.LineGeometry对象无法进行自我绘制,因此同样需要使用 Path元素来辅助呈现.LineGeometry ...

  7. Java面向对象程序设计的六大基本原则

    1.开闭原则(Open Close Principle) 定义:一个软件实体如类.模块和函数应该对扩展开放,对修改关闭. 开放-封闭原则的意思就是说,你设计的时候,时刻要考虑,尽量让这个类是足够好,写 ...

  8. 如何判断java对象是否为String数组

    if (entry.getValue() instanceof String[]) {// ko .................... }

  9. MySQL 实现将一个库表里面的数据实时更新到另一个库表里面

    MySQL 实现将一个库表里面的数据实时更新到另一个库表里面 需求描述:MySQL 里面有很多的数据库,这些数据库里面都有同一种表结构的表 (tb_warn_log),这张表的数据是实时更新的,现在需 ...

  10. ThinkPHP 3.2 性能优化,实现高性能API开发

    需求分析 目前的业务全站使用ThinkPHP 3.2.3,前台.后台.Cli.Api等.目前的业务API访问量数千万,后端7台PHP 5.6,平均CPU使用率20%. 测试数据 真实业务 php5.6 ...