#include <bits/stdc++.h>
using namespace std;

#define rep(_, __, ___) for (int _ = __; _ <= ___; ++_)
#define int long long

int f[1005][1005];
int rev[1005][1005];
int n;
int op[1005], a[1005];
int res[1005], maxans;

signed main()
{
	scanf("%lld", &n);
	rep(i, 1, n)
	{
		char ch = getchar();
		while (!isalpha(ch))
			ch = getchar();
		if (ch == 't')
			op[i] = op[n + i] = 1;
		else
			op[i] = op[n + i] = 2;
		scanf("%lld", a + i);
		a[n + i] = a[i];	
	}
	rep(i, 1, n)
	{
		int ln = i, rn = n + i - 1;
		memset(f, -0x3f, sizeof(f));
		memset(rev, 0x3f, sizeof(rev));
		rep(j, ln, rn)
				f[j][j] = rev[j][j] = a[j];
		int ans = -1073741824;
		rep(jn, 1, n - 1)
			rep(kn, ln, rn - jn)
			{
				int j = kn, k = kn + jn;
				rep(l, j, k - 1)
				{
					if (op[l + 1] == 1)
					{
						f[j][k] = max(f[j][k], f[j][l] + f[l + 1][k]);
//						f[j][k] = max(f[j][k], rev[j][l] + rev[l + 1][k]);
						rev[j][k] = min(rev[j][k], f[j][l] + f[l + 1][k]);
//						rev[j][k] = min(rev[j][k], rev[j][l] + rev[l + 1][k]);
					}
					else
					{
						f[j][k] = max(f[j][k], f[j][l] * f[l + 1][k]);
						f[j][k] = max(f[j][k], rev[j][l] * rev[l + 1][k]);
						f[j][k] = max(f[j][k], rev[j][l] * f[l + 1][k]);
						f[j][k] = max(f[j][k], f[j][l] * rev[l + 1][k]);
						rev[j][k] = min(rev[j][k], f[j][l] * f[l + 1][k]);
						rev[j][k] = min(rev[j][k], rev[j][l] * rev[l + 1][k]);
						rev[j][k] = min(rev[j][k], rev[j][l] * f[l + 1][k]);
						rev[j][k] = min(rev[j][k], f[j][l] * rev[l + 1][k]);
					}
				}
			}
		ans = max(ans, f[ln][rn]);
		maxans = max(ans, maxans);
		res[i] = ans;
	}
//	int var1 = res[1];
//	rep(i, 1, n - 1)
//		res[i] = res[i + 1];
//	res[n] = var1;
	printf("%lld\n", maxans);
	rep(i, 1, n)
		if (res[i] == maxans)
			printf("%lld ", i);
	printf("\n");
	rep(i, 1, n)
		cerr << res[i] << ' ';
	return 0;
}

$\texttt{WA on 5\#}$

思路大概和第一篇题解是一样的，数组已经开大，已经全局 $\texttt{long long}$，大概看了一下转移方程也差不多，实在不知道怎么改了。

求调。