贪心搞就行,用map记录每个数出现的下标,每次都取首尾两个。将中间权值为负的删掉后取sum值最大的就行。

#include<iostream>
#include<algorithm>
#include<vector>
#include<string>
#include<stack>
#include<map>
#include<set>
#include<cstdio>
#include<cstring>
#include<cstdio>
#include<cmath>
#include<cstdlib>
#define FF(i, a, b) for(int i=a; i<b; i++)
#define FD(i, a, b) for(int i=a; i>b; i--)
#define REP(i, n) for(int i=0; i<n; i++)
#define CLR(a, b) memset(a, b, sizeof(a))
#define LL long long using namespace std; const int maxn = 333333;
int n, a[maxn], l, r, k, ans[maxn];
LL s[maxn];
map<int, vector<int> > mp;
map<int, vector<int> > :: iterator it; int main()
{
while(~scanf("%d", &n))
{
s[0] = 0; mp.clear();
LL sum = -111111111111, tmp;
FF(i, 1, 1+n)
{
scanf("%d", &a[i]);
if(a[i] > 0) s[i] = s[i-1] + a[i];
else s[i] = s[i-1];
mp[a[i]].push_back(i);
}
for(it = mp.begin(); it != mp.end(); it++)
{
int nc = it->second.size();
if(nc >= 2)
{
tmp = s[it->second[nc-1]] - s[it->second[0]-1];
if(it->first < 0) tmp += it->first*2;
if(tmp > sum)
{
sum = tmp, l = it->second[0], r = it->second[nc-1];
}
}
}
k = 0;
FF(i, 1, l) ans[k++] = i;
FF(i, l+1, r) if(a[i] < 0) ans[k++] = i;
FF(i, r+1, n+1) ans[k++] = i;
printf("%I64d %d\n", sum, k);
REP(i, k) printf("%d ", ans[i]);
}
return 0;
}

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