CodeForces Gym 100685I Innovative Business (贪心)

目录

作者：@dwtfukgv
本文为作者原创，转载请注明出处：https://www.cnblogs.com/dwtfukgv/p/5825774.html

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题意：给定一条路的长和宽，然后给你瓷砖的长和宽，你只能横着或者竖着铺，也可以切成片，但是每条边只能对应一条边，问你最少要多少瓷砖。

析：先整块整块的放，然后再考虑剩下部分，剩下的再分成3部分，先横着，再竖着，最后是他们相交的部分。

代码如下：

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#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 frer freopen("in.txt", "r", stdin) #define frew freopen("out.txt", "w", stdout) using namespace std;   typedef long long LL; typedef pair<int, int> P; const int INF = 0x3f3f3f3f; const double inf = 0x3f3f3f3f3f3f; const double PI = acos(-1.0); const double eps = 1e-8; const int maxn = 1e3 + 5; const int mod = 1e9 + 7; const int dr[] = {-1, 0, 1, 0}; const int dc[] = {0, 1, 0, -1}; const char *de[] = {"0000", "0001", "0010", "0011", "0100", "0101", "0110", "0111", "1000", "1001", "1010", "1011", "1100", "1101", "1110", "1111"}; 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(){     int tw, tl, rw, rl;     while(scanf("%d %d %d %d", &rl, &rw, &tl, &tw) == 4){         int ans = rl/tl * (rw/tw);         int l = rl % tl;         int w = rw % tw;           if(l == 0 && w == 0)  printf("%d\n", ans);         else if(l == 0 && w != 0){             int x = tw / w;             ans += rl/tl/x;             ans += (rl/tl) % x != 0;             printf("%d\n", ans);         }         else if(l != 0 && w == 0){             int x = tl / l;             ans += rw/tw/x;             ans += (rw/tw) % x != 0;             printf("%d\n", ans);         }         else{             int x = tw / w;             ans += rl/tl/x;             ans += (rl/tl) % x != 0;             int y = tl / l;             ans += rw/tw/y;             ans += (rw/tw) % y != 0;             if((rl/tl) % x == 0 && (rw/tw) % y == 0)  ++ans;             printf("%d\n", ans);         }     }     return 0; }