杭电多校 hdu6627 equation

http://acm.hdu.edu.cn/showproblem.php?pid=6627

题意：解绝对值方程并统计解的个数。

解法：签到题，直接模拟小学数学学的零点分段法即可。（数据多直接cin,cout会T，还以为是算法有问题...

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const int maxn = 1e5+10;
ll s1[maxn],s2[maxn];

struct node{
    ll a,b;
    double s;
}g[maxn];

bool cmp(node p,node q){
    return p.s>q.s;
}

struct note{
    ll x,y;
    double s;
}v[maxn];

bool cmp1(note p1,note q1){
    return p1.s<q1.s;
}

map<double,int> ma;

int main(){
    ios::sync_with_stdio(0);
    cin.tie(0);
    cout.tie(0);
    int t;
    cin>>t;
    ll n,c;
    while(t--){
        int cnt = 0;
        ma.clear();
        cin>>n>>c;
        for(ll i=1;i<=n;i++){
            cin>>g[i].a>>g[i].b;
            g[i].s = -1.0*g[i].b/g[i].a;
        }
        sort(g+1,g+n+1,cmp);
        s1[0] = 0; s2[0] = 0;
        for(ll i=1;i<=n;i++){
            s1[i] = s1[i-1]+g[i].a;
            s2[i] = s2[i-1]+g[i].b;
        }
        int flag = 0;
        for(ll i=0;i<=n;i++){
            ll a1 = s1[n]-2*s1[i];
            ll a2 = s2[n]-2*s2[i];
            ll s = c-a2;            //a1*x=s
            if(a1==0&&s==0){
                flag = 1;
                break;
            }
            else if(a1==0){
                continue;
            }
            else {
                ll num1 = abs(s);
                ll num2 = abs(a1);
                ll gcd = __gcd(num1,num2);
                double ss = (double)s/a1;
                if((i==n||ss>g[i+1].s)&&(i==0||ss<=g[i].s)){
                    if(ma[ss]==0){
                        ma[ss]++;
                        v[cnt].x = s/gcd;
                        v[cnt].y = a1/gcd;
                        v[cnt].s = ss;
                        cnt++;
                    }
                }
            }
        }
        if(flag==1){
            cout<<-1<<endl;
        }
        else{
            cout<<cnt;
            sort(v,v+cnt,cmp1);
            for(ll i=0;i<cnt;i++){
                if(v[i].x*v[i].y>0){
                    cout<<' '<<abs(v[i].x)<<'/'<<abs(v[i].y);
                }
                else if(v[i].x*v[i].y==0){
                    cout<<' '<<"0/1";
                }
                else {
                    cout<<" -"<<abs(v[i].x)<<'/'<<abs(v[i].y);
                }
            }
            cout<<endl;
        }
    }
    return 0;
}