码迷,mamicode.com
首页 > 其他好文 > 详细

Sum It Up 广搜

时间:2014-07-16 19:11:38      阅读:236      评论:0      收藏:0      [点我收藏+]

标签:des   style   blog   http   color   strong   

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

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
 
 
bubuko.com,布布扣
 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <map>
 4 #include <string.h>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 using namespace std;
 9 int t,a[20],ans[20],n,nu;
10 void dfs(int s,int as,int sum)
11 {
12     int i;
13     if(sum==t)
14     {
15         nu++;
16         for(i=0;i<as;i++)
17         if(i!=as-1)
18         printf("%d+",ans[i]);
19         else printf("%d\n",ans[i]);
20         return ;
21     }
22     if(s>=n)return ;
23     for(i=s;i<n;i++)
24     {
25         ans[as]=a[i];
26         dfs(i+1,as+1,sum+a[i]);
27         while(i+1<n&&a[i]==a[i+1])i++;
28     }
29 }
30 int main()
31 {
32     //freopen("in.txt","r",stdin);
33     int i,j;
34     while(scanf("%d%d",&t,&n),t||n)
35     {
36         nu=0;
37         for(i=0;i<n;i++)
38         scanf("%d",&a[i]);
39         cout<<"Sums of "<<t<<":"<<endl;
40         dfs(0,0,0);
41         if(nu==0)
42         cout<<"NONE"<<endl;
43     }
44 }
View Code

 

bubuko.com,布布扣
 1 #include <iostream>
 2 #include <stdio.h>
 3 #include <map>
 4 #include <string.h>
 5 #include <algorithm>
 6 #include <queue>
 7 #include <vector>
 8 using namespace std;
 9 int t,n,a[20][2],ans[20][2],an,ok;
10 void dfs(int s,int anss,int sum)
11 {
12     //cout<<s<<" "<<anss<<" "<<sum<<endl;
13     int i,j;
14     if(sum>t)return;
15     if(sum==t)
16     {
17         for(i=0; i<anss; i++)
18         {
19             for(j=0; j<ans[i][1]; j++)
20             if(i==anss-1&&j==ans[i][1]-1)
21             printf("%d\n",ans[i][0]);
22             else
23                 printf("%d+",ans[i][0]);
24         }
25         ok++;
26         return ;
27     }
28     if(s>=an)return;
29     for(j=a[s][1]; j>=0; j--)
30     {
31         ans[anss][0]=a[s][0];
32         ans[anss][1]=j;
33         if(j)
34             dfs(s+1,anss+1,sum+a[s][0]*j);
35         else dfs(s+1,anss,sum+a[s][0]*j);
36     }
37     return ;
38 }
39 int main()
40 {
41     //freopen("in.txt","r",stdin);
42     int i,j,x;
43     while(scanf("%d%d",&t,&n),t||n)
44     {
45         ok=0;
46         an=0;
47         for(i=0; i<n; i++)
48         {
49             scanf("%d",&x);
50             if(an==0)
51             {
52                 a[an][0]=x,a[an++][1]=1;
53             }
54             else
55             {
56                 if(x==a[an-1][0])
57                     a[an-1][1]++;
58                 else  a[an][0]=x,a[an++][1]=1;
59             }
60         }
61         cout<<"Sums of "<<t<<":"<<endl;
62         dfs(0,0,0);
63         if(!ok)
64         cout<<"NONE"<<endl;
65     }
66 }
View Code

 

 

Sum It Up 广搜,布布扣,bubuko.com

Sum It Up 广搜

标签:des   style   blog   http   color   strong   

原文地址:http://www.cnblogs.com/ERKE/p/3844658.html

(0)
(0)
   
举报
评论 一句话评论(0
登录后才能评论!
© 2014 mamicode.com 版权所有  联系我们:gaon5@hotmail.com
迷上了代码!