CodeForces Gym 100685I Innovative Business (贪心)

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：

#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
", ans);
        else if(l == 0 && w != 0){
            int x = tw / w;
            ans += rl/tl/x;
            ans += (rl/tl) % x != 0;
            printf("%d
", ans);
        }
        else if(l != 0 && w == 0){
            int x = tl / l;
            ans += rw/tw/x;
            ans += (rw/tw) % x != 0;
            printf("%d
", 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
", ans);
        }
    }
    return 0;
}