前天晚上没做 昨晚上回去补了两道题
A题：
有两条地铁线路 一条正向1, 2, 3, ..., n 另一条逆向n, n-1, n-2, ..., 1 有两个人分别沿正向和逆向出发 给出总站数 还有两人的起点和终点 问两人是否能相遇
模拟+简单数学题啦
AC代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned long long ull; 5 6 namespace io { 7 const int SIZE = 1e7 + 10; 8 char inbuff[SIZE]; 9 char *l, *r; 10 inline void init() { 11 l = inbuff; 12 r = inbuff + fread(inbuff, 1, SIZE, stdin); 13 } 14 inline char gc() { 15 if (l == r) init(); 16 return (l != r) ? *(l++) : EOF; 17 } 18 void read(int &x) { 19 x = 0; char ch = gc(); 20 while(!isdigit(ch)) ch = gc(); 21 while(isdigit(ch)) x = x * 10 + ch - '0', ch = gc(); 22 } 23 } using io::read; 24 25 int main(){ 26 int n, a, x, b, y; 27 cin>>n>>a>>x>>b>>y; 28 bool flag = false; 29 for (int i = 0;; i++){ 30 if (a == b) flag = true; 31 a++; 32 b--; 33 if (a == n + 1) a = 1; 34 if (b == 0) b = n; 35 if (a == x || b == y) break; 36 } 37 if (a == b) flag = true; 38 if (flag) cout<<"YES"<<endl; 39 else cout<<"NO"<<endl; 40 return 0; 41 }
B题 给出m对在1~n之间的数 现在问 是否能找出两个不相等的数x y 使得每一对数至少有一个数等于x或y
假设一对数(a, b)中至少有一个数为x或y 下一对(c, d)中 若(c, d) != (a, b) 则(c, d)中也至少存在一个数为x或y 即(x, y)为(a, c) || (a, d) || (b, c) || (b, d) 就只需要判断这m对数是否满足这个条件就行啦 若(c, d) == (a, b) 那（a, b)中就一个数为x 另一个数为y
AC代码：
1 #include<bits/stdc++.h> 2 using namespace std; 3 typedef long long ll; 4 typedef unsigned long long ull; 5 6 namespace io { 7 const int SIZE = 1e7 + 10; 8 char inbuff[SIZE]; 9 char *l, *r; 10 inline void init() { 11 l = inbuff; 12 r = inbuff + fread(inbuff, 1, SIZE, stdin); 13 } 14 inline char gc() { 15 if (l == r) init(); 16 return (l != r) ? *(l++) : EOF; 17 } 18 void read(int &x) { 19 x = 0; char ch = gc(); 20 while(!isdigit(ch)) ch = gc(); 21 while(isdigit(ch)) x = x * 10 + ch - '0', ch = gc(); 22 } 23 } using io::read; 24 25 pair<int, int> p[300005]; 26 int n, m; 27 28 bool judge(int a, int b){ 29 for (int i = 0; i < m; i++) 30 if (p[i].first != a && p[i].first != b && 31 p[i].second != a && p[i].second != b) 32 return false; 33 return true; 34 } 35 36 int main(){ 37 cin>>n>>m; 38 int a, b; 39 int c = 0, d = 0; 40 bool flag = false; 41 cin>>p[0].first>>p[0].second; 42 a = p[0].first; 43 b = p[0].second; 44 for (int i = 1; i < m; i++){ 45 cin>>p[i].first>>p[i].second; 46 if (p[i].first != a && p[i].first != b && 47 p[i].second != a && p[i].second != b){ 48 c = p[i].first; 49 d = p[i].second; 50 } 51 } 52 if (judge(a, b) || judge(a, c) || judge(a, d) || judge(b, c) || judge(b, d)) 53 flag = true; 54 if (flag) cout<<"YES"<<endl; 55 else cout<<"NO"<<endl; 56 return 0; 57 }