【题意】一个n*m的矩形广场，用a*a的方形石板铺设，问最少需要多少块石板能铺满广场。
【思路】水题，从n方向来看能能够铺设ceil(n/a)块，从m方向来看能能够铺设ceil(m/a)块，总共有ceil(n/a)*ceil(m/a)块。
1 /*
2 ** CodeForces 1A Theatre Square
3 ** Created by Rayn @@ 2014/05/18
4 */
5 #include <cstdio>
6 #include <cmath>
7 typedef long long LL;
8
9 int main()
10 {
11 #ifdef _Rayn
12 freopen("in.txt", "r", stdin);
13 #endif
14
15 int n, m, a;
16 scanf("%d%d%d", &n, &m, &a);
17 //LL l1 = (n + a - 1) / a; //另一种写法，很巧妙
18 //LL l2 = (m + a - 1) / a;
19 LL l1 = n / a;
20 if (n % a != 0)
21 l1++;
22 LL l2 = m / a;
23 if (m % a != 0)
24 l2++;
25 //printf("%I64d %I64d\n", l1, l2);
26 LL ans = l1 * l2;
27 printf("%I64d\n", ans);
28 return 0;
29 }
【题意】题意好复杂，难得写了，就是对于表格的单元表示有两种表示法，一种是RXCY，代表X行Y列，另一种是字母加数字，字母是代表列，数字代表行，A-Z代表1到26，
AA代表27，AZ代表52列，AAA代表703列。依次类推。要求你对输入的格式转化为另一种格式。
【思路】开始没有想明白到底怎么转化，各种情况处理，if
else，while满天飞，WA了好多次，后来简化来看这个就是类似于26进制的道理，然后代码就很清晰了。
1 /*
2 ** CodeForces 1B Spreadsheets
3 ** Created by Rayn @@ 2014/05/19
4 */
5 #include <cstdio>
6 #include <cmath>
7 #include <cstring>
8 #include <cctype>
9 typedef long long LL;
10 const int MAX = 100010;
11
12 char coor[MAX], col[MAX];
13
14 void GetRow(char *str, int &r, int &c)
15 {
16 int flag = 1;
17 r = c = 0;
18 for (int i = 1; i < strlen(str); ++i)
19 {
20 if (isdigit(str[i]) && flag)
21 {
22 r = r * 10 + str[i] - ‘0‘;
23 }
24 if (isdigit(str[i]) && !flag)
25 {
26 c = c * 10 + str[i] - ‘0‘;
27 }
28 if (str[i] == ‘C‘)
29 {
30 flag = 0;
31 }
32 }
33 }
34 void GetCol(char *str, char *col, int &r)
35 {
36 int cnt = 0;
37 r = 0;
38 for (int i = 0; i < strlen(str); ++i)
39 {
40 if (isalpha(str[i]))
41 {
42 col[cnt++] = str[i];
43 }
44 if (isdigit(str[i]))
45 {
46 r = r * 10 + str[i] - ‘0‘;
47 }
48 }
49 }
50 int main()
51 {
52 #ifdef _Rayn
53 freopen("in.txt", "r", stdin);
54 #endif
55
56 int t;
57 scanf("%d%*c", &t);
58 while (t--)
59 {
60 memset(col, 0, sizeof(col));
61 gets(coor);
62 int type = 0;
63 if (coor[0] == ‘R‘ && isdigit(coor[1]))
64 {
65 for (int i = 2; i < strlen(coor); ++i)
66 {
67 if (coor[i] == ‘C‘)
68 {
69 type = 1;
70 break;
71 }
72 }
73 }
74 if (type == 1)
75 {
76 int rown = 0, coln = 0, cnt = 0;
77 GetRow(coor, rown, coln);
78 while (coln)
79 {
80 coln--;
81 int t1 = coln % 26;
82 col[cnt++] = t1 + ‘A‘;
83 coln /= 26;
84 }
85 for (int i = cnt - 1; i >= 0; --i)
86 {
87 printf("%c", col[i]);
88 }
89 printf("%d\n", rown);
90 }
91 else if (type == 0)
92 {
93 int rown = 0, coln = 0;
94 GetCol(coor, col, rown);
95 coln = 0;
96 for (int i = 0; i < strlen(col); ++i)
97 {
98 coln = coln * 26 + (col[i] - ‘A‘);
99 coln++;
100 //printf("%d\n", coln);
101 }
102 printf("R%dC%d\n",rown, coln);
103 }
104 }
105 return 0;
106 }
107 /*
108 ** CodeForces 1B Spreadsheets
109 ** Created by Rayn @@ 2014/05/19
110 */
111 #include <cstdio>
112 #include <cmath>
113 #include <cstring>
114 #include <cctype>
115 typedef long long LL;
116 const int MAX = 100010;
117
118 char coor[MAX], col[MAX];
119
120 void GetRow(char *str, int &r, int &c)
121 {
122 int flag = 1;
123 r = c = 0;
124 for (int i = 1; i < strlen(str); ++i)
125 {
126 if (isdigit(str[i]) && flag)
127 {
128 r = r * 10 + str[i] - ‘0‘;
129 }
130 if (isdigit(str[i]) && !flag)
131 {
132 c = c * 10 + str[i] - ‘0‘;
133 }
134 if (str[i] == ‘C‘)
135 {
136 flag = 0;
137 }
138 }
139 }
140 void GetCol(char *str, char *col, int &r)
141 {
142 int cnt = 0;
143 r = 0;
144 for (int i = 0; i < strlen(str); ++i)
145 {
146 if (isalpha(str[i]))
147 {
148 col[cnt++] = str[i];
149 }
150 if (isdigit(str[i]))
151 {
152 r = r * 10 + str[i] - ‘0‘;
153 }
154 }
155 }
156 int main()
157 {
158 #ifdef _Rayn
159 freopen("in.txt", "r", stdin);
160 #endif
161
162 int t;
163 scanf("%d%*c", &t);
164 while (t--)
165 {
166 memset(col, 0, sizeof(col));
167 gets(coor);
168 int type = 0;
169 if (coor[0] == ‘R‘ && isdigit(coor[1]))
170 {
171 for (int i = 2; i < strlen(coor); ++i)
172 {
173 if (coor[i] == ‘C‘)
174 {
175 type = 1;
176 break;
177 }
178 }
179 }
180 if (type == 1)
181 {
182 int rown = 0, coln = 0, cnt = 0;
183 GetRow(coor, rown, coln);
184 while (coln)
185 {
186 coln--;
187 int t1 = coln % 26;
188 col[cnt++] = t1 + ‘A‘;
189 coln /= 26;
190 }
191 for (int i = cnt - 1; i >= 0; --i)
192 {
193 printf("%c", col[i]);
194 }
195 printf("%d\n", rown);
196 }
197 else if (type == 0)
198 {
199 int rown = 0, coln = 0;
200 GetCol(coor, col, rown);
201 coln = 0;
202 for (int i = 0; i < strlen(col); ++i)
203 {
204 coln = coln * 26 + (col[i] - ‘A‘);
205 coln++;
206 //printf("%d\n", coln);
207 }
208 printf("R%dC%d\n",rown, coln);
209 }
210 }
211 return 0;
212 }
【题意】旧时罗马广场的的斗兽场形似一个等边等角的凸包多边形，每个顶点都有一个柱子。由于年代久远，只剩下3个柱子，由此构造一个多边形，使面积最小。最多只会出现100边的凸包。
【思路】浓浓的几何题，各种公式的推导花了我一节高数课的时间。【待补完。。。】
1 /*
2 ** CodeForces 1C Ancient Berland Circus
3 ** Created by Rayn @@ 2014/05/19
4 */
5 #include <cstdio>
6 #include <cmath>
7 #include <cstring>
8 #define EPS 1e-5
9 typedef long long LL;
10 const int MAX = 1010;
11 const double PI = acos(-1.0);
12
13 struct Pillar
14 {
15 double x, y;
16 } point[5];
17
18 double Length(Pillar a, Pillar b)
19 {
20 return sqrt((b.x - a.x)*(b.x - a.x) + (b.y - a.y)*(b.y - a.y));
21 }
22 double GetS(double a, double b, double c)
23 {
24 double p = (a + b + c) / 2;
25 return sqrt(p*(p - a)*(p - b)*(p - c));
26 }
27 double fgcd(double x, double y)
28 {
29 while (fabs(x) > EPS && fabs(y) > EPS)
30 {
31 if (x > y)
32 x -= floor(x / y)*y;
33 else
34 y -= floor(y / x)*x;
35 }
36 return x + y;
37 }
38 int main()
39 {
40 #ifdef _Rayn
41 freopen("in.txt", "r", stdin);
42 #endif
43 double la, lb, lc, S, R;
44 double angA, angB, angC;
45
46 for (int i = 0; i < 3; ++i)
47 {
48 scanf("%lf%lf", &point[i].x, &point[i].y);
49 }
50 la = Length(point[0], point[1]);
51 lb = Length(point[0], point[2]);
52 lc = Length(point[1], point[2]);
53 S = GetS(la, lb, lc);
54 R = la * lb * lc / (4 * S);
55
56 angA = acos((lb*lb + lc*lc - la*la) / (2 * lb * lc));
57 angB = acos((la*la + lc*lc - lb*lb) / (2 * la * lc));
58 angC = acos((la*la + lb*lb - lc*lc) / (2 * la * lb));
59 double n = PI / fgcd(fgcd(angA, angB), angC);
60 double ans = n/2 * sin(2 * PI / n) * R*R;
61 printf("%.6f\n", ans);
62
63 return 0;
64 }
CodeForces Beta Round #1,布布扣,bubuko.com
时间： 2025-01-01 05:00:47