题干：
You are given an array aa consisting of nn integers. You can perform the following operations with it:
- Choose some positions ii and jj (1≤i,j≤n,i≠j1≤i,j≤n,i≠j), write the value of ai⋅ajai⋅aj into the jj-th cell and remove the number from the ii-th cell;
- Choose some position ii and remove the number from the ii-th cell (this operation can be performed no more than once and at any point of time, not necessarily in the beginning).
The number of elements decreases by one after each operation. However, the indexing of positions stays the same. Deleted numbers can't be used in the later operations.
Your task is to perform exactly n−1n−1 operations with the array in such a way that the only number that remains in the array is maximum possible. This number can be rather large, so instead of printing it you need to print any sequence of operations which leads to this maximum number. Read the output format to understand what exactly you need to print.
Input
The first line contains a single integer nn (2≤n≤2⋅1052≤n≤2⋅105) — the number of elements in the array.
The second line contains nn integers a1,a2,…,ana1,a2,…,an (−109≤ai≤109−109≤ai≤109) — the elements of the array.
Output
Print n−1n−1 lines. The kk-th line should contain one of the two possible operations.
The operation of the first type should look like this: 1 ik jk1 ik jk, where 11 is the type of operation, ikik and jkjk are the positions of the chosen elements.
The operation of the second type should look like this: 2 ik2 ik, where 22 is the type of operation, ikik is the position of the chosen element. Note that there should be no more than one such operation.
If there are multiple possible sequences of operations leading to the maximum number — print any of them.
Examples
Input
5
5 -2 0 1 -3
Output
2 3
1 1 2
1 2 4
1 4 5
Input
5
5 2 0 4 0
Output
1 3 5
2 5
1 1 2
1 2 4
Input
2
2 -1
Output
2 2
Input
4
0 -10 0 0
Output
1 1 2
1 2 3
1 3 4
Input
4
0 0 0 0
Output
1 1 2
1 2 3
1 3 4
Note
Let X be the removed number in the array. Let's take a look at all the examples:
The first example has, for example, the following sequence of transformations of the array: [5,−2,0,1,−3]→[5,−2,X,1,−3]→[X,−10,X,1,−3]→[5,−2,0,1,−3]→[5,−2,X,1,−3]→[X,−10,X,1,−3]→ [X,X,X,−10,−3]→[X,X,X,X,30][X,X,X,−10,−3]→[X,X,X,X,30]. Thus, the maximum answer is 3030. Note, that other sequences that lead to the answer 3030 are also correct.
The second example has, for example, the following sequence of transformations of the array: [5,2,0,4,0]→[5,2,X,4,0]→[5,2,X,4,X]→[X,10,X,4,X]→[5,2,0,4,0]→[5,2,X,4,0]→[5,2,X,4,X]→[X,10,X,4,X]→ [X,X,X,40,X][X,X,X,40,X]. The following answer is also allowed:
1 5 3
1 4 2
1 2 1
2 3
Then the sequence of transformations of the array will look like this: [5,2,0,4,0]→[5,2,0,4,X]→[5,8,0,X,X]→[40,X,0,X,X]→[5,2,0,4,0]→[5,2,0,4,X]→[5,8,0,X,X]→[40,X,0,X,X]→ [40,X,X,X,X][40,X,X,X,X].
The third example can have the following sequence of transformations of the array: [2,−1]→[2,X][2,−1]→[2,X].
The fourth example can have the following sequence of transformations of the array: [0,−10,0,0]→[X,0,0,0]→[X,X,0,0]→[X,X,X,0][0,−10,0,0]→[X,0,0,0]→[X,X,0,0]→[X,X,X,0].
The fifth example can have the following sequence of transformations of the array: [0,0,0,0]→[X,0,0,0]→[X,X,0,0]→[X,X,X,0][0,0,0,0]→[X,0,0,0]→[X,X,0,0]→[X,X,X,0].
题目大意：
n个数，两种操作，第一种是a[i]*a[j],删掉a[i],第二种是直接删除a[i]（其中第二种操作只能用一次）剩下的数序列号不变。操作n-1次，使最后剩下的那个数最大化。
让你输出这n-1次操作（数字位置的序号输出原序列的）。
解题报告：
分情况，如果全是正数或者有偶数个负数，那就全乘起来，然后0单独处理。
如果有奇数个负数，那就贪心将绝对值最小的那个负数看成0一类的，然后化为第一种情况，一样处理。
样例给的很到位啊，，，不然就忘了判断那个操作次数是否是n-1次了，因为只有1操作不到n-1次时，才用2操作。
再就是x刚开始给了-1e9，，，竟然WA了，，不过想想也是啊，，这个值还是太大了，，赋初值-1e9-1也行、、
AC代码：
#include<cstdio>
#include<iostream>
#include<algorithm>
#include<queue>
#include<map>
#include<vector>
#include<set>
#include<string>
#include<cmath>
#include<cstring>
#define F first
#define S second
#define ll long long
#define pb push_back
#define pm make_pair
using namespace std;
typedef pair<int,int> PII;
const int MAX = 2e5 + 5;
int pos[MAX],a[MAX];
int tot,cnt;
int main()
{int n,cur=0;cin>>n;for(int i = 1; i<=n; i++) {scanf("%d",a+i);if(a[i] == 0) pos[++tot] = i;if(a[i] < 0) cnt++;}if(cnt%2==0) {int last = -1;for(int i = 1; i<=n; i++) {if(a[i] == 0) continue;else {if(last == -1) last = i;else printf("1 %d %d\n",last,i),last=i,cur++;}}for(int i = 2; i<=tot; i++) printf("1 %d %d\n",pos[i-1],pos[i]),cur++;if(cur!=n-1) printf("2 %d\n",pos[tot]);}else {int x=-1e9-12,tar;for(int i = 1; i<=n; i++) {if(a[i] >= 0) continue;if(a[i] > x) tar=i,x=a[i];}tot=0;for(int i = 1; i<=n; i++) {if(i==tar||a[i]==0) pos[++tot]=i;}int last = -1;for(int i = 1; i<=n; i++) {if(a[i] == 0 || i == tar) continue;else {if(last == -1) last = i;else printf("1 %d %d\n",last,i),last=i,cur++;}}for(int i = 2; i<=tot; i++) printf("1 %d %d\n",pos[i-1],pos[i]),cur++;if(cur!=n-1) printf("2 %d\n",pos[tot]); }return 0 ;
}
)
