目录
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:
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 30 31 32 33 34 35 36 37 38 39 40 41 42 43 44 45 46 47 48 49 50 51 52 53 54 55 56 57 58 59 60 61 62 63 64 65 66 67 68 69 70 71 72 73 74 | #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; } |
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