HNU Digit Sum （状态压缩）

Digit Sum

Time Limit: 2000ms, Special Time Limit:5000ms, Memory Limit:65536KB

Total submit users: 69, Accepted users: 61

Problem 12944 : No special judgement

Problem description

When Grace was in third grade, her elementary school teacher assigned her the following problem: What is the smallest possible sum of two numbers that together use the numerals 1, 2, 7, 8, and 9? Grace figured out that the answer to this problem is 207 (for example, as 78 + 129), but when the teacher assigned four pages of similar problems as homework, Grace got bored. It turns out that Grace was a rather advanced third grader, so she decided that it would be more fun to write a computer program to solve such problems. Surely you can do the same!

Input

Input:  Each problem is described on a single line. The line begins with an integer N, such that 2 ≤ N ≤ 14, designating the number of numerals included in the problem. Following that are those N numerals. There will always be at least 2 numerals that are nonzero. The end of the input is designated by a line containing only the value 0.

Output

Output:  For each case, output a line with the minimum sum S that can be achieved. Please keep in mind that by standard convention, the numeral 0 cannot appear as the first digit of either summand.

Sample Input

5 1 2 7 8 9 6 3 4 2 2 2 2 9 0 1 2 3 4 0 1 2 3 0

Sample Output

207 447 11257

Problem Source

ACM Mid-Central Programming Competition 2013 #include #include using namespace std; #define mulit(i) (1<<(i)) __int64 a[15],sta[mulit(14)]; void init(int n) { __int64 sum; int h,i; sort(a,a+n); for(int s=1;s0) { h=i; sum=a[i]; break; } if(mulit(i-1)==s) { sta[s]=a[i-1]; continue; } if(mulit(i)>s) { sta[s]=-1; continue; } for( i=0;mulit(i)<=s;i++) if((mulit(i)&s)&&h!=i) sum=sum*10+a[i]; sta[s]=sum; } } int main() { int n; __int64 minsum; while(scanf("%d",&n)>0&&n) { for(int i=0;ista[s]+sta[s^(mulit(n)-1)]) minsum=sta[s]+sta[s^(mulit(n)-1)]; printf("%I64d\n",minsum); } }