首页 > 专题系列 > Java解POJ > POJ 1564 Sum It Up [解题报告] Java
2013
11-09

POJ 1564 Sum It Up [解题报告] Java

Sum It Up

问题描述 :

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.

输入:

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.

输出:

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.

样例输入:

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

样例输出:

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

解题代码:

/* @author: */
import java.util.Scanner;
public class Main{
  static int sum,n;
  static int arr[]=new int[20];
  static int j,FLAG;
  static boolean used[]=new boolean[20];

  static void f(int now,int total)
  {
   int i;
   if(total==sum)
   {
     FLAG=1;
     for(i=0;i< n;i++)
     if(used[i])
     {
	System.out.printf("%d",arr[i]);
	  break;
      }
     for(i++;i< n;i++)
     {
	if(used[i])
	  System.out.printf("+%d",arr[i]);
      }
      System.out.println("");
	return;
     }
     for(i=now;i< n;i++)
     {
	if(total+arr[i]>sum) continue;
	if(i>0&&arr[i]==arr[i-1]&&!used[i-1]) continue;
	used[i]=true;
	f(i+1,total+arr[i]);
	used[i]=false;
      }
   }
public static void main(String args[])
{
   Scanner sc=new Scanner(System.in);
   while(true)
   {
    sum=sc.nextInt();
    n=sc.nextInt();
   if(n==0) break;
    FLAG=0;
   for(j=0;j< n;j++)
     arr[j]=sc.nextInt();
   System.out.printf("Sums of %d:\n",sum);
   f(0,0);
   if(FLAG==0) System.out.println("NONE");
   }
  }
}

  1. 样例输出和程序输出不吻合,修改一下样例输出吧。我用的是VC编译器,会提示我的i和j变量重复定义

  2. #!/usr/bin/env python
    def cou(n):
    arr =
    i = 1
    while(i<n):
    arr.append(arr[i-1]+selfcount(i))
    i+=1
    return arr[n-1]

    def selfcount(n):
    count = 0
    while(n):
    if n%10 == 1:
    count += 1
    n /= 10
    return count

  3. 题目需要求解的是最小值,而且没有考虑可能存在环,比如
    0 0 0 0 0
    1 1 1 1 0
    1 0 0 0 0
    1 0 1 0 1
    1 0 0 0 0
    会陷入死循环