【HDOJ6627】equation（模拟）

题意：给定n，整数序列a和b，整数C，求所有成立的x

n<=1e5，1<=a[i]<=1e3，-1e3<=b[i]<=1e3，1<=C<=1e9

思路：

大概就照每条直线的零点分段，维护一下系数和常数项

特判的地方挺多，精度也要注意，写起来像计算几何

感觉这种麻烦的东西应该有模板

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef unsigned int uint;
typedef unsigned long long ull;
typedef pair<int,int> PII;
typedef pair<ll,ll> Pll;
typedef vector<int> VI;
typedef vector<PII> VII;
#define N  1100000
#define M  4100000
#define fi first
#define se second
#define MP make_pair
#define pi acos(-1)
#define mem(a,b) memset(a,b,sizeof(a))
#define rep(i,a,b) for(int i=(int)a;i<=(int)b;i++)
#define per(i,a,b) for(int i=(int)a;i>=(int)b;i--)
#define lowbit(x) x&(-x)
#define Rand (rand()*(1<<16)+rand())
#define id(x) ((x)<=B?(x):m-n/(x)+1)
#define ls p<<1
#define rs p<<1|1

const ll MOD=998244353,inv2=(MOD+1)/2;
      double eps=1e-6;
      int INF=1e9;

struct arr
{
    ll a,b;
}c[N],ans[N];

bool cmp(arr a,arr b)
{
    return a.b*b.a>b.b*a.a;
}

ll read()
{
   ll v=0,f=1;
   char c=getchar();
   while(c<48||57<c) {if(c=='-') f=-1; c=getchar();}
   while(48<=c&&c<=57) v=(v<<3)+v+v+c-48,c=getchar();
   return v*f;
}

ll gcd(ll x,ll y)
{
    if(y==0) return x;
    return gcd(y,x%y);
}

int xiaoyu(ll x1,ll y1,ll x2,ll y2)
{
    //printf("xiaoyu %I64d %I64d %I64d %I64d
",x1,y1,x2,y2);
    int p=-1;
    ll t=x1*y2-y1*x2;
    if(t==0) return 1;
    if(t<0) p=-p;
    if(y1*y2<0) p=-p;
    return (p==1);
}

int main()
{
    //freopen("1.in","r",stdin);
    int cas=read();
    while(cas--)
    {
        int n;
        scanf("%d",&n);
        ll C=read();
        rep(i,1,n)
        {
            c[i].a=read();
            c[i].b=read();
        }
        sort(c+1,c+n+1,cmp);
        //rep(i,1,n) printf("%I64d %I64d
",c[i].a,c[i].b);
        int m=0;
        ll sa=0,sb=0;
        rep(i,1,n)
        {
            sa-=c[i].a;
            sb-=c[i].b;
        }
        //printf("sa=%I64d C-sb=%I64d
",sa,C-sb);
        int flag=0;
        if(sa==0&&C-sb==0) flag=1;
        if(xiaoyu(C-sb,sa,-c[1].b,c[1].a))
        {
            if(sa==0&&C-sb!=0) continue;
            m++;
            ll t=gcd(abs(C-sb),abs(sa));
            //printf("t=%I64d
",t);
            ans[m].a=(C-sb)/t;
            ans[m].b=sa/t;
            if(ans[m].b<0)
            {
                ans[m].a=-ans[m].a;
                ans[m].b=-ans[m].b;
            }
        }
        c[n+1].a=-1e15; c[n+1].b=1;
        rep(i,1,n)
        {
            sa+=2ll*c[i].a;
            sb+=2ll*c[i].b;
            //printf("sa=%I64d C-sb=%I64d
",sa,C-sb);
            if(sa==0&&C-sb==0)
            {
                flag=1;
                break;
            }
            if(sa==0&&C-sb!=0) continue;
            if(xiaoyu(C-sb,sa,-c[i].b,c[i].a)==0&&xiaoyu(C-sb,sa,-c[i+1].b,c[i+1].a))
            {
                if(m>=1&&(C-sb)*ans[m].b==sa*ans[m].a) continue;
                m++;
                ll t=gcd(abs(C-sb),abs(sa));
                ans[m].a=(C-sb)/t;
                ans[m].b=sa/t;
                if(ans[m].b<0)
                {
                    ans[m].a=-ans[m].a;
                    ans[m].b=-ans[m].b;
                }
            }
        }
        if(flag)
        {
            printf("-1
");
            continue;
        }
        if(m==0)
        {
            printf("0
");
            continue;
        }

        printf("%d ",m);
        rep(i,1,m-1) printf("%I64d/%I64d ",ans[i].a,ans[i].b);
        printf("%I64d/%I64d
",ans[m].a,ans[m].b);
    }

    return 0;
}