题意：https://codeforc.es/problemset/problem/195/D
求折线段数。
思路：
对pos进行sort，对不同区间段加k，两个dp处理不同k>0 or k<0前后缀，判断即可。
注意：long double，ESP=1e-20。
1 #define IOS ios_base::sync_with_stdio(0); cin.tie(0); 2 #include <cstdio>//sprintf islower isupper 3 #include <cstdlib>//malloc exit strcat itoa system("cls") 4 #include <iostream>//pair 5 #include <fstream>//freopen("C:\Users\13606\Desktop\草稿.txt","r",stdin); 6 #include <bitset> 7 //#include <map> 8 //#include<unordered_map> 9 #include <vector> 10 #include <stack> 11 #include <set> 12 #include <string.h>//strstr substr 13 #include <string> 14 #include <time.h>//srand(((unsigned)time(NULL))); Seed n=rand()%10 - 0~9; 15 #include <cmath> 16 #include <deque> 17 #include <queue>//priority_queue<int, vector<int>, greater<int> > q;//less 18 #include <vector>//emplace_back 19 //#include <math.h> 20 //#include <windows.h>//reverse(a,a+len);// ~ ! ~ ! floor 21 #include <algorithm>//sort + unique : sz=unique(b+1,b+n+1)-(b+1);+nth_element(first, nth, last, compare) 22 using namespace std;//next_permutation(a+1,a+1+n);//prev_permutation 23 #define fo(a,b,c) for(register int a=b;a<=c;++a) 24 #define fr(a,b,c) for(register int a=b;a>=c;--a) 25 #define mem(a,b) memset(a,b,sizeof(a)) 26 #define pr printf 27 #define sc scanf 28 #define ls rt<<1 29 #define rs rt<<1|1 30 typedef long long ll; 31 #define RG register int; 32 void swapp(int &a,int &b); 33 double fabss(double a); 34 int maxx(int a,int b); 35 int minn(int a,int b); 36 int Del_bit_1(int n); 37 int lowbit(int n); 38 int abss(int a); 39 //const long long INF=(1LL<<60); 40 const double E=2.718281828; 41 const double PI=acos(-1.0); 42 const int inf=(1<<30); 43 const double ESP=1e-20; 44 const int mod=(int)1e9+7; 45 const int N=(int)1e6+10; 46 47 struct node 48 { 49 int k,b,id; 50 long double pos; 51 friend bool operator<(node a,node b) 52 { 53 return a.pos<b.pos; 54 } 55 }a[N]; 56 57 long double get(int k,int b) 58 { 59 long double len=1.0*b/(1.0*k); 60 len=abs(len); 61 if(k>0) 62 { 63 if(b>0) 64 return -len; 65 else 66 return len; 67 } 68 else 69 { 70 if(b>0) 71 return len; 72 else 73 return -len; 74 } 75 } 76 bool same(long double x,long double y) 77 { 78 return abs(x-y)<ESP; 79 } 80 81 ll dp[N],dp2[N]; 82 83 int main() 84 { 85 int n,cnt=0; 86 long double xx; 87 sc("%d",&n); 88 for(int i=1;i<=n;++i) 89 { 90 int k,b; 91 sc("%d%d",&k,&b); 92 if(k==0) 93 continue; 94 a[++cnt]={k,b},a[cnt].pos=get(a[cnt].k,a[cnt].b); 95 } 96 n=cnt; 97 sort(a+1,a+1+n); 98 a[1].id=1; 99 int id=1; 100 for(int i=2;i<=n;++i) 101 { 102 if(!same(a[i].pos,a[i-1].pos)) 103 id++; 104 a[i].id=id; 105 } 106 for(int i=1;i<=n;++i) 107 if(a[i].k>0) 108 dp[a[i].id]+=a[i].k; 109 for(int i=1;i<=n;++i) 110 dp[i]+=dp[i-1]; 111 for(int i=n;i>=1;--i) 112 { 113 if(a[i].k<0) 114 dp2[a[i].id-1]+=a[i].k; 115 } 116 for(int i=n;i>=0;--i) 117 dp2[i]+=dp2[i+1]; 118 int ans=0; 119 for(int i=1;i<=id;++i) 120 if(dp[i]+dp2[i]!=dp[i-1]+dp2[i-1]) 121 ans++; 122 pr("%d ",ans); 123 return 0; 124 } 125 126 /**************************************************************************************/ 127 128 int maxx(int a,int b) 129 { 130 return a>b?a:b; 131 } 132 133 void swapp(int &a,int &b) 134 { 135 a^=b^=a^=b; 136 } 137 138 int lowbit(int n) 139 { 140 return n&(-n); 141 } 142 143 int Del_bit_1(int n) 144 { 145 return n&(n-1); 146 } 147 148 int abss(int a) 149 { 150 return a>0?a:-a; 151 } 152 153 double fabss(double a) 154 { 155 return a>0?a:-a; 156 } 157 158 int minn(int a,int b) 159 { 160 return a<b?a:b; 161 }