2A Subset——折半枚举+二分

题目

Given a list of N integers with absolute values no larger than (10^15), find a non empty subset of these numbers which minimizes the absolute value of the sum of its elements. In case there are multiple subsets, choose the one with fewer elements.

Input

The input contains multiple data sets, the first line of each data set contains N <= 35, the number of elements, the next line contains N numbers no larger than (10^15) in absolute value and separated by a single space. The input is terminated with N = 0

Output

For each data set in the input print two integers, the minimum absolute sum and the number of elements in the optimal subset.

Sample Input

1
10
3
20 100 -100
0

Sample Output

10 1
0 2

题解

解题思路

首先想到暴力，看到n<=35，TLE无疑了
而如果看一半，18还是可以的
所以这道题就是将这些数分成前后两半，分别计算
算后面的时候在前面二分查找相反数最接近的数进行计算
这道题用int会会爆掉，要用到long long
但这样abs就不能用了，得自己手写一个

代码

#include <cstdio>
#include <cmath>
#include <map>
#define int long long
using namespace std;
const int N = 1e3+5;
int n, a[N];
int mabs(int x) {
    return x > 0 ? x : -x;
}
signed main() {
    while (1) {
        scanf("%lld", &n);
        if (!n) break;
        for(int i = 1; i <= n; i++)
            scanf("%lld", &a[i]);
        int half = n >> 1;
        pair<int, int> ans(mabs(a[1]), 1);
        map<int, int> m;
        map<int, int>::iterator it;
        for(int s = 1; s < (1 << half); s++) {
            int sum = 0, cnt = 0;
            for(int i = 1; i <= half; i++)
                if (s & (1 << (i - 1))) sum += a[i], cnt++;
            pair<int, int> x(mabs(sum), cnt);
            ans = min(ans, x);
            if (m[sum]) m[sum] = min(m[sum], cnt);
            else m[sum] = cnt;
        }
        for(int s = 1; s < (1 << (n - half)); s++) {
            int sum = 0, cnt = 0;
            for(int i = 1; i <= n - half; i++)
                if (s & (1 << (i - 1))) sum += a[i+half], cnt++;
            pair<int, int> x(mabs(sum), cnt);
            ans = min(ans, x);
            it = m.lower_bound(-sum);
            if (it != m.end()) {
                pair<int, int> x(mabs(sum + it->first), cnt + it->second);
                ans = min(ans, x);
                
            }
            if (it != m.begin()) {
                it--;
                pair<int, int> x(mabs(sum + it->first), cnt + it->second);
                ans = min(ans, x);
            }
        }
        printf("%lld %lld
", ans.first, ans.second);
    }
    return 0;
}