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Gym 100531G Grave(水题)
题意:给定一个大矩形,再给定在一个小矩形,然后给定一个新矩形的长和高,问你能不能把这个新矩形放到大矩形里,并且不与小矩形相交。
析:直接判定小矩形的上下左右四个方向,能不能即可。
代码如下:
#pragma comment(linker, "/STACK:1024000000,1024000000") #include <cstdio> #include <string> #include <cstdlib> #include <cmath> #include <iostream> #include <cstring> #include <set> #include <queue> #include <algorithm> #include <vector> #include <map> #include <cctype> #include <cmath> #include <stack> #define debug puts("+++++") //#include <tr1/unordered_map> #define freopenr freopen("in.txt", "r", stdin) #define freopenw freopen("out.txt", "w", stdout) using namespace std; //using namespace std :: tr1; typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const LL LNF = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 50000 + 5; const LL mod = 1e3 + 7; const int N = 1e6 + 5; const int dr[] = {-1, 0, 1, 0, 1, 1, -1, -1}; const int dc[] = {0, 1, 0, -1, 1, -1, 1, -1}; const char *Hex[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; inline LL gcd(LL a, LL b){ return b == 0 ? a : gcd(b, a%b); } inline int gcd(int a, int b){ return b == 0 ? a : gcd(b, a%b); } inline int lcm(int a, int b){ return a * b / gcd(a, b); } int n, m; const int mon[] = {0, 31, 28, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; const int monn[] = {0, 31, 29, 31, 30, 31, 30, 31, 31, 30, 31, 30, 31}; inline int Min(int a, int b){ return a < b ? a : b; } inline int Max(int a, int b){ return a > b ? a : b; } inline LL Min(LL a, LL b){ return a < b ? a : b; } inline LL Max(LL a, LL b){ return a > b ? a : b; } inline bool is_in(int r, int c){ return r >= 0 && r < n && c >= 0 && c < m; } int main(){ freopen("grave.in", "r", stdin); freopen("grave.out", "w", stdout); int x1, y1, x2, y2, x3, y3, w, h, x4, y4; while(cin >> x1 >> y1 >> x2 >> y2){ cin >> x3 >> y3 >> x4 >> y4; cin >> w >> h; bool ok = false; if(y2 - y4 >= h && x2 - x1 >= w) ok = true; else if(y3 - y1 >= h && x2 - x1 >= w) ok = true; else if(x3 - x1 >= w && y2 - y1 >= h) ok = true; else if(x2 - x4 >= w && y2 - y1 >= h) ok = true; printf("%s\n", ok ? "Yes" : "No"); } return 0; }
Gym 100531G Grave(水题)
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