[洛谷P3878][TJOI2010]分金币

题目大意：把$n(n\leqslant30)$个数分成两组，两组个数最多相差$1$，求出两组元素差的绝对值最小使多少

题解：模拟退火

卡点：$\exp$中的两个数相减写反，导致$\exp(x)$中的$x>0$，$\exp(x)>1$，相当于一直接受生成的解

C++ Code：

#include <algorithm>
#include <cstdio>
#include <cmath>
#define maxn 32
inline long long abs(long long a) {return a > 0 ? a : -a;}
int T, n, divn;
long long sum;
 
const double ST = 100, delT = 0.9992, eps = 1e-5;
int Tim = 20;
struct node {
	int s[maxn];
	long long w;
} ans;
inline long long calc(node &x) {
	long long __sum = 0;
	for (int i = 0; i < divn; i++) __sum += x.s[i];
	x.w = abs(sum - __sum - __sum);
	if (n & 1) x.w = std::min(x.w, abs(sum - __sum - __sum - x.s[divn] - x.s[divn]));
	return x.w;
}
inline double rand_d() {return static_cast<double> (rand()) / RAND_MAX;}
 
void SA() {
	double T = ST;
	node now = ans, nxt;
	while (T > eps) {
		nxt = now;
		int x = rand() % n, y = rand() % n;
		T *= delT;
		if (x == y) continue;
		std::iter_swap(nxt.s + x, nxt.s + y);
		long long del = calc(nxt);
		if (del < now.w || exp((now.w - del) / T) > rand_d()) now = nxt;
		if (del < ans.w) ans = nxt;
	}
}
int main() {
	srand(20040826);
	scanf("%d", &T);
	while (T --> 0) {
		scanf("%d", &n); divn = n >> 1;
		sum = 0;
		for (int i = 0; i < n; i++) scanf("%d", ans.s + i), sum += ans.s[i];
		if (n == 1) {
			printf("%d\n", *ans.s);
			continue;
		}
		std::random_shuffle(ans.s, ans.s + n);
		calc(ans);
		for (int i = 0; i < Tim; i++) SA();
		printf("%lld\n", ans.w);
	}
	return 0;
}