CF1042C Array Product 分类讨论+贪心

考虑有无负数（负数的个数为奇视作“有”，否则为“无”）和有无零

无负数无零，全部合并即可

无负数有零，那么把零合并起来，删掉零

有负数无零，把最大的负数找出来，删掉，合并剩余的数

有负数有零，把零和最大的负数合并起来，删掉，合并剩余的数

注意如果只剩下一个数，不能删掉这唯一的一个数

#include <vector>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
namespace remoon {
    #define ri register int
    #define ll long long
    #define tpr template <typename ra>
    #define rep(iu, st, ed) for(ri iu = st; iu <= ed; iu ++)
    #define drep(iu, ed, st) for(ri iu = ed; iu >= st; iu --)
    #define gc getchar
    inline int read() {
        int p = , w = ; char c = gc();
        while(c > '' || c < '') { if(c == '-') w = -; c = gc(); }
        while(c >= '' && c <= '') p = p *  + c - '', c = gc();
        return p * w;
    }
    int wr[], rw;
    #define pc(iw) putchar(iw)
    tpr inline void write(ra o, char c = '\n') {
        if(!o) pc('');
        if(o < ) o = -o, pc('-');
        while(o) wr[++ rw] = o % , o /= ;
        while(rw) pc(wr[rw --] + '');
        pc(c);
    }
    tpr inline void cmin(ra &a, ra b) { if(a > b) a = b; }
    tpr inline void cmax(ra &a, ra b) { if(a < b) a = b; }
    tpr inline bool ckmin(ra &a, ra b) { return (a > b) ? a = b,  : ; }
    tpr inline bool ckmax(ra &a, ra b) { return (a < b) ? a = b,  : ; }
}
using namespace std;
using namespace remoon;
#define sid 300050

int n, a[sid];

inline void Solve1() {
    rep(i, , n) printf("1 %d %d\n", i, );
}

inline void Solve2() {

    int mip = -1e9 - , pos = ;
    rep(i, , n)
    if(a[i] <  && ckmax(mip, a[i])) pos = i;

    int fir = -;
    rep(i, , n) if(pos != i) { fir = i; break; }
    if(fir == -) return;

    printf("2 %d\n", pos);
    if(pos != ) {
        rep(i, , n)
        if(pos != i) printf("%d %d %d\n", , i, );
    }
    else rep(i, , n) printf("%d %d %d\n", , i, );

}

inline void Solve3() {

    int lst = -;
    rep(i, , n) if(a[i] == ) {
        if(lst != -) printf("%d %d %d\n", , lst, i);
           lst = i;
    }

    int fir = -;
    rep(i, , n) if(a[i] != ) { fir = i; break; }
    if(fir == -) return;

    printf("2 %d\n", lst);
    rep(i, fir + , n) if(a[i] != )
    printf("%d %d %d\n", , i, fir);

}

inline void Solve4() {

    int mip = -1e9 - , pos = ;
    rep(i, , n)
    if(a[i] <  && ckmax(mip, a[i])) pos = i;

    int lst = -;
    rep(i, , n) if(a[i] ==  || pos == i) {
        if(lst != -)  printf("%d %d %d\n", , lst, i);
        lst = i;
    }

    int fir = -;
    rep(i, , n) if(a[i] !=  && pos != i)
    { fir = i; break; }
    if(fir == -) return;

    printf("2 %d\n", lst);
    rep(i, fir + , n) if(a[i] !=  && pos != i)
    printf("%d %d %d\n", , i, fir);

}

int main() {

    n = read();
    rep(i, , n) a[i] = read();

    int neg = , zero = ;
    rep(i, , n) if(a[i] < ) neg ^= ;
    else if(a[i] == ) zero |= ;

    if(!zero && !neg) Solve1();
    else if(!zero && neg) Solve2();
    else if(zero && !neg) Solve3();
    else if(zero && neg) Solve4();

    return ;
}