Sum It Up
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 3538 Accepted Submission(s):
1788
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.
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.
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.
'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
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
这是一道搜索题，题目的大意是给你一个数t，用所给的n个数找出所有和为t的种类，注意1+2和2+1是同一种。首先应该把数组按从大到小的顺序排序，然后进行搜索，如果和sum等于t，输出结果。还要注意剪纸的时候，要避免输出重复的答案。即需要在dfs()调用函数之后找到a[i+1]不等于a[i]的i(
while(a[i+1] == a[i]) i++;)。
#include <iostream>
#include <algorithm>
using namespace std;
int t,n;
int a[],b[];
bool flag;
bool cmp(int a,int b)
{
return a>b;
}
void dfs(int sum,int k,int j) //k代表要求得数组下标,j代表整个数组的下标
{
int i;
if (sum>t)
return;
if (sum == t)
{
cout<<b[];
for (i=;i<k;i++)
cout<<"+"<<b[i];
cout<<endl;
flag = false;
return ;
}
for(i=j;i<n;i++)
{
if(sum+a[i]>t)
continue;
b[k] = a[i];
dfs(sum+a[i],k+,i+);
while(a[i+] == a[i]) //剪纸的重要一步，找到下一个不相等的数
i++;
}
}
int main()
{
int i;
while(cin>>t>>n && t && n)
{
flag = true;
for(i=;i<n;i++)
cin>>a[i];
sort(a,a+n,cmp);
cout<<"Sums of "<<t<<":"<<endl;
dfs(,,);
if (flag)
cout<<"NONE"<<endl;
}
return ;
}