hdu 1258

Sum It Up

Time Limit: 2000/1000 MS (Java/Others)    Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 6448    Accepted Submission(s):
3365

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

 自己拿set判的重，代码还有瑕疵，数据太水水过。哎

这道题也是每次两种决策，加上这个数或者不加，判重类似于全排列的判重方式。

#include <iostream>
#include <cstdio>
#include <cstring>
#include <algorithm>
using namespace std;
int num,n,pos;
int a[15],b[15];
bool judge = false;
void output(int depth)
{
for(int i =0 ;i< depth; ++i)
if(!i) printf("%d",b[i]);
else printf("+%d",b[i]);
printf("\n");
}
void dfs(int depth,int sum,int pos)                                //每一位数字只有两种可能（加/不加），pos表示当前进行到了哪一位，depth表示b里保存的数字数量便于打印
{
if(sum == num) {judge = true;output(depth); return;}
if(sum>num) return;// 超出了 终止递归
if(pos>=n) return; //选择的数的位置超出数据范围
b[depth] = a[pos];
dfs(depth+1,sum+a[pos],pos+1);
while(pos+1<n&&a[pos] == a[pos+1]) pos++;//关键  判重
dfs(depth,sum,pos+1);

}
int main()
{

while(scanf("%d%d",&num,&n) && num){
printf("Sums of %d:\n",num);
for(int i = 0; i<n; ++i) scanf("%d",&a[i]);
judge = false;
dfs(0,0,0);

if(judge == false) printf("NONE\n");
}
return 0;
}