• 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;
    }
    
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  • 原文地址:https://www.cnblogs.com/dwtfukgv/p/5825774.html
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