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Codeforces Round #308 (Div. 2)

分类: IT文章 • 2023-11-25 11:43:49

Codeforces Round #308 (Div. 2)

AB 水题

C 进制转换，挺有意思

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <fstream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <deque>
 7 #include <vector>
 8 #include <queue>
 9 #include <string>
10 #include <cstring>
11 #include <map>
12 #include <stack>
13 #include <set>
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 //#define OPEN_FILE
17 using namespace std;
18 vector<int> p;
19 int main()
20 {
21 #ifdef OPEN_FILE
22     freopen("in.txt", "r", stdin);
23     freopen("out.txt", "w", stdout);
24 #endif // OPEN_FILE
25     int w, m;
26     p.empty();
27     scanf("%d%d", &w, &m);
28     bool flag;
29     while (m != 0){
30         p.push_back(m % w);
31         m /= w;
32     }
33     flag = false;
34     for (int i = 0; i < p.size(); i++){
35         if (p[i] == 0 || p[i] == 1){
36             continue;
37         }
38         if (p[i] - w == -1){
39             p[i] = -1;
40             if (p.size() - i == 1){
41                 break;
42             }
43             p[i + 1] += 1;
44             int j = i + 1;
45             while (p[j] >= w){
46                 if (p.size() - j == 1){
47                     p[j] = 0;
48                     p.push_back(1);
49                     break;
50                 }
51                 p[j] = 0;
52                 p[j + 1] += 1;
53                 j++;
54             }
55         }
56         else{
57             flag = true;
58             break;
59         }
60     }
61     if (flag){
62         printf("NO
");
63     }
64     else{
65         printf("YES
");
66     }
67 }
View Code

D 平面中N个点，找出三点不共线的数目

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <fstream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <deque>
 7 #include <vector>
 8 #include <queue>
 9 #include <string>
10 #include <cstring>
11 #include <map>
12 #include <stack>
13 #include <set>
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 #define MAXM 2001
17 using namespace std;
18 int x[MAXM], y[MAXM];
19 double a[MAXM];
20 bool compare(double a, double b){
21     return a < b;
22 }
23 int main()
24 {
25     int n;
26     scanf("%d", &n);
27     for (int i = 1; i <= n; i++){
28         scanf("%d%d", &x[i], &y[i]);
29     }
30     LL ans;
31     ans = (n - 1) * (n - 2);
32     ans = ans * n / 6;
33     int m = 0;
34     for (int i = 1; i <= n; i++){
35         m = 0;
36         for (int j = i + 1; j <= n; j++){
37             if (x[j] == x[i]){
38                 a[++m] = INF;
39                 continue;
40             }
41             double temp1 = y[j] - y[i];
42             double temp2 = x[j] - x[i];
43             a[++m] = temp1 / temp2;
44             if (a[m] == 1.0){
45                 a[m] = a[m];
46             }
47         }
48         sort(a + 1, a + m + 1);
49         int t = 1;
50         for (int j = 2; j <= m; j++){
51             if (a[j] - a[j - 1] < 1e-7){
52                 ++t;
53                 continue;
54             }
55             ans -= t *(t - 1) / 2;
56             t = 1;
57         }
58         if (t != 1){
59             ans -= t *(t - 1) / 2;
60         }
61     }
62     printf("%I64d
", ans);
63 }
View Code

E 表达式计算，乘号很少，枚举乘号位置即可

 1 #include <iostream>
 2 #include <cstdio>
 3 #include <fstream>
 4 #include <algorithm>
 5 #include <cmath>
 6 #include <deque>
 7 #include <vector>
 8 #include <queue>
 9 #include <string>
10 #include <cstring>
11 #include <map>
12 #include <stack>
13 #include <set>
14 #define LL long long
15 #define INF 0x3f3f3f3f
16 //#define OPEN_FILE
17 using namespace std;
18 char s[5005];
19 int len;
20 LL calc(int a, int b){
21     LL x, y;
22     x = 0;
23     y = s[a + 1] - '0';
24     for (int i = a + 2; i <= b - 2; i += 2){
25         if (s[i] == '+'){
26             x += y;
27             y = s[i + 1] - '0';
28         }
29         else{
30             y *= s[i + 1] - '0';
31         }
32     }
33     LL temp = x + y;
34     x = 0;
35     y = s[1] - '0';
36     for (int i = 2; i <= len; i += 2){
37         if (i == a){
38             y *= temp;
39             i = b;
40         }
41         if (s[i] == '+'){
42             x += y;
43             y = s[i + 1] - '0';
44         }
45         else{
46             y *= s[i + 1] - '0';
47         }
48     }
49     return x + y;
50 }
51 int main()
52 {
53 #ifdef OPEN_FILE
54     freopen("in.txt", "r", stdin);
55     freopen("out.txt", "w", stdout);
56 #endif // OPEN_FILE
57     scanf("%s", s + 3);
58     s[1] = '1';
59     s[2] = '*';
60     len = strlen(s + 1);
61     s[len + 1] = '*';
62     s[len + 2] = '1';
63     len += 2;
64     LL ans = -1;
65     for(int i = 2; i <= len; i += 2){
66         for (int j = i + 2; j <= len; j += 2){
67             if (s[i] == '*' && s[j] == '*'){
68                 ans = max(ans, calc(i, j));
69             }
70         }
71     }
72     printf("%I64d
", ans);
73 }
View Code

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