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 ;

}