讨论+细节 c题

codeforces 1042 c

思路简单，但代码较复杂的贪心
分类讨论：

• 有0
  • 负数有奇数个：将最大的负数和0全部乘到一起，最后删掉0
  • 负数有偶数个：将0全部乘到一起，最后删掉0
• 没有0
  • 负数有奇数个：将最大绝对值最小的负数删掉
  • 负数有偶数个：不删

     

#include <iostream>
#include<string.h>
#include<cstdio>
#include<algorithm>
using namespace std;
typedef long long ll;
int main()
{
    ll n,i,j,k,t,maxx=-0x3f3f3f3f,t1,a[200010]={0},fu=0,ling=0,c[200010],d[200010];
    scanf("%lld",&n);
    for(i=0; i<=n-1; i++)
    {
        scanf("%lld",&a[i]);
        if(a[i]<0)
        {
            fu++;
            if(a[i]>maxx)
            {
                maxx=a[i];
                k=i;
            }
        }
        if(a[i]==0)
            ling++;
        
    }
    if(ling!=0)
    {
        if(fu%2==0)
        {
            memset(c,0,sizeof(c));
            memset(d,0,sizeof(d));
            t=0;
            t1=0;
            int ge=0;
            for(i=0; i<=n-1; i++)
            {
                if(a[i]==0)
                    c[t++]=i+1;
                if(a[i]!=0)
                    d[t1++]=i+1;
            }
            for(i=0; i<=t-2; i++)
            {
                printf("1 %lld %lld\n",c[i],c[i+1]);
                ge++;
            }
            if(ge==n-1)
                return 0;
            printf("2 %lld\n",c[t-1]);
            for(i=0; i<=t1-2; i++)
            {
                printf("1 %lld %lld\n",d[i],d[i+1]);

            }
        }
        else
        {
            memset(c,0,sizeof(c));
            memset(d,0,sizeof(d));
            t=0;
            t1=0;
            int ge=0;
            for(i=0; i<=n-1; i++)
            {
                if(a[i]==0||i==k)
                    c[t++]=i+1;
                else
                    d[t1++]=i+1;
            }
            for(i=0; i<=t-2; i++)
            {
                printf("1 %lld %lld\n",c[i],c[i+1]);
                ge++;
            }
            if(ge==n-1)
                return 0;
            printf("2 %lld\n",c[t-1]);
            for(i=0; i<=t1-2; i++)
            {
                printf("1 %lld %lld\n",d[i],d[i+1]);

            }

        }
    }
    else
    {
        if(fu%2!=0)
        {
            memset(c,0,sizeof(c));
            memset(d,0,sizeof(d));
            t=0;
            t1=0;
            int ge=0;
            for(i=0; i<=n-1; i++)
            {
                if(i==k)
                    c[t++]=i+1;
                else
                    d[t1++]=i+1;
            }
            for(i=0; i<=t-2; i++)
            {
                
                printf("1 %lld %lld\n",c[i],c[i+1]);
                ge++;
            }
            if(ge==n-1)
                return 0;
            printf("2 %lld\n",c[t-1]);
            for(i=0; i<=t1-2; i++)
            {
                printf("1 %lld %lld\n",d[i],d[i+1]);

            }
        }
        else
        {
            memset(c,0,sizeof(c));
            memset(d,0,sizeof(d));
            t=0;
            t1=0;
            for(i=0; i<=n-1; i++)
            {
                if(a[i]!=0)
                    d[t1++]=i+1;
            }
            for(i=0; i<=t1-2; i++)
            {
                printf("1 %lld %lld\n",d[i],d[i+1]);
            }
        }

    }


}