A. Fafa and his Company
方法：暴力枚举leader的个数即可
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27
28 int main()
29 {
30 int n;
31 scanf("%d", &n);
32 int ans = 0;
33 for (int i = 1; i < n; ++i)
34 if ((n-i)%i == 0)
35 ++ans;
36 printf("%d\n", ans);
37 }
B. Fafa and the gates
方法：暴力模拟
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27
28 int n;
29 const int maxn = 1e5+1;
30 char s[maxn];
31 int get(int x, int y)
32 {
33 if (x == y)
34 return 0;
35 if (x > y)
36 return 1;
37 else
38 return -1;
39 }
40
41 int main()
42 {
43 scanf("%d %s", &n, s);
44 int cur = 0, ans = 0;
45 int x = 0, y = 0;
46 for (int i = 0; i < n; ++i)
47 {
48 if (s[i] == ‘U‘)
49 ++y;
50 else
51 ++x;
52 int tmp = get(x, y);
53 if (tmp)
54 {
55 if (tmp == -cur)
56 {
57 ++ans;
58 }
59 cur = tmp;
60 }
61 }
62 printf("%d\n", ans);
63 }
C. Fifa and Fafa
题意：给你一个圆和一个点，让你在给定圆内画一个圆，使得答案圆不能包含给定点，而且使得给定圆没有被答案圆覆盖的面积最小。给出答案圆的圆心和半径即可。
观察：给定圆没有被覆盖的面积最小，即使答案圆面积最大。如果给定点在给定圆之外，那么答案圆等于给定圆。如果给定点在给定圆之内，只需特判一下给定点和圆心重合的情况，剩下的情况，答案圆可以被唯一确定，即作给定点到给定圆圆心的射线，交给定圆与一交点。以交点和给定点之间线段为直径的圆就是答案。
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27
28 ll r, x1, yy1, x2, y2;
29 ll dis()
30 {
31 ll dx = x1-x2, dy = yy1-y2;
32 return dx*dx + dy*dy;
33 }
34 void end(double x, double y, double r)
35 {
36 printf("%.6lf %.6lf %.6lf\n", x, y, r);
37 exit(0);
38 }
39 int main()
40 {
41 scanf("%lld %lld %lld %lld %lld", &r, &x1, &yy1, &x2, &y2);
42 if (dis() >= r*r)
43 end(x1, yy1, r);
44 if (dis() == 0)
45 end(x1+r/2.0, yy1, r/2.0);
46 double R = (sqrt(dis())+r)/2.0;
47 double dx = (x1-x2), dy = (yy1-y2);
48 dx *= R/sqrt(dis()), dy *= R/sqrt(dis());
49 end(x2+dx, y2+dy, R);
50
51
52 }
D. Fafa and Ancient Alphabet
题意：给你两个长度为n的字符串a, b，字母表大小是m，1 <= n, m <= 1e5。字符串中有一些字符是问号。让你把问好替换成字母表中的数字，问你有多少种方法可以是第一个字符串的字典序大于第二个。
观察：一位一位的讨论即可。
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27 const int mod = 1e9+7;
28 ll power(ll base, ll p)
29 {
30 ll ret = 1;
31 while (p)
32 {
33 if (p&1)
34 ret = ret * base % mod;
35 base = base * base % mod;
36 p >>= 1;
37 }
38 return ret;
39 }
40 const int maxn = 1e5+1;
41 int n, m;
42 ll p[maxn<<1];
43 int a[2][maxn];
44 int main()
45 {
46 scanf("%d %d", &n, &m);
47 int tot = 0;
48 for (int i = 0; i < 2; ++i)
49 for (int j = 0; j < n; ++j)
50 scanf("%d", &a[i][j]), tot += !a[i][j];
51 p[0] = 1;
52 for (int i = 1; i < maxn*2; ++i)
53 p[i] = p[i-1] * m % mod;
54 ll ans = 0;
55 ll base = 1;
56 ll cnt = 0;
57 for (int i = 0; i < n; ++i)
58 {
59 cnt += !a[0][i];
60 cnt += !a[1][i];
61 if (!a[0][i] && !a[1][i])
62 {
63 ans = (ans + base * p[tot-cnt] % mod * (1ll * m * (m-1) / 2 % mod) % mod) % mod;
64 base = base * m % mod;
65 }
66 else if (!a[0][i])
67 {
68 ans = (ans + base * p[tot-cnt] % mod * (m-a[1][i]) % mod) % mod;
69 }
70 else if (!a[1][i])
71 {
72 ans = (ans + base * p[tot-cnt] % mod * (a[0][i]-1) % mod) % mod;
73 }
74 else if (a[0][i] > a[1][i])
75 {
76 ans = (ans + base * p[tot-cnt] % mod) % mod;
77 break;
78 }
79 else if (a[0][i] < a[1][i])
80 {
81 break;
82 }
83 }
84 ans = ans * power(p[tot], mod-2) % mod;
85 printf("%lld\n", ans);
86 }
E. Fafa and Ancient Mathematics
题意：定义合法表达式：一个单数位的正整数（即1-9）是一个合法的表达式。对于任意两个表达式E1和E2，(E1+E2), (E1-E2)都是合法的表达式。下面给你一个长度不超过1e4的表达式，其中所有运算符号（即+和-）都被替换成了？。又告诉你这些问号中，p个是+，剩下的n个是-，min(p, n) <= 100。让你输出表达式最大可能的值。
观察：可以发现表达式是一颗二叉树。很容易想到一个dp，maxi[cur][cnt+][cnt-]（(mini[cur][cnt+][cnt-]）表示在cur表达式中，或者说cur表达式对应的子树中，填cnt+个+和cnt-个-，能得到的最大值（最小值）。因为题目保证min(p, n) <= 100，而且是二叉树，所以可以把总数量较小的符号记入状态，比如p <= n, maxi[cur][cnt] /mini[cur][cnt]，表示cur子树中填cnt个加号能得到的最大值/最小值。计算maxi[cur][cnt]/mini[cur][cnt] 只需要枚举左右子树内的加号个数，和中间符号是+还是-就好了。
code：
1 /*
2 by skydog
3 */
4 #include <iostream>
5 #include <cstdio>
6 #include <vector>
7 #include <utility>
8 #include <algorithm>
9 #include <cmath>
10 #include <cstring>
11 #include <map>
12 #include <set>
13 #include <stack>
14 #include <queue>
15 #include <deque>
16 #include <cassert>
17 #include <list>
18 using namespace std;
19 typedef long long ll;
20 typedef pair<int, int> ii;
21 typedef pair<ll, ll> l4;
22
23 #define mp make_pair
24 #define pb push_back
25 #define db(x) cerr << #x << " = " << x << endl
26
27 const int maxn = 1e4+1;
28 char s[maxn];
29 int sz = 0, ptr = 0;
30 vector<int> g[maxn];
31 const int maxp = 101;
32 int maxi[maxn][maxp];
33 int mini[maxn][maxp];
34 int v[maxn];
35 int p, m;
36 bool isp;
37 inline void Max(int &a, int b)
38 {
39 if (a < b) a = b;
40 }
41 inline void Min(int &a, int b)
42 {
43 if (a > b) a = b;
44 }
45 int dfs(int cur)
46 {
47 fill(maxi[cur], maxi[cur]+maxp, -1e5);
48 fill(mini[cur], mini[cur]+maxp, 1e5);
49 if (isdigit(s[ptr]))
50 {
51 v[cur] = s[ptr]-‘0‘;
52 ++ptr;
53 maxi[cur][0] = mini[cur][0] = v[cur];
54 return 0;
55 }
56 assert(s[ptr] == ‘(‘);
57 ++ptr;
58 int lhs, rhs, ret = 1;
59 g[cur].pb(++sz);
60 ret += dfs(lhs = sz);
61 assert(s[ptr] == ‘?‘);
62 ++ptr;
63 g[cur].pb(++sz);
64 ret += dfs(rhs = sz);
65 assert(s[ptr] == ‘)‘);
66 ++ptr;
67 if (isp)
68 {
69 //limited positive
70 for (int tot = 1; tot < maxp; ++tot)
71 {
72 //pick mid to be positive
73 for (int i = 0; i <= tot-1; ++i)
74 {
75 Max(maxi[cur][tot], maxi[lhs][i]+maxi[rhs][tot-1-i]);
76 Min(mini[cur][tot], mini[lhs][i]+mini[rhs][tot-1-i]);
77 }
78 }
79 for (int tot = 0; tot < maxp; ++tot)
80 {
81 //pick mid to be negative
82 for (int i = 0; i <= tot; ++i)
83 {
84 Max(maxi[cur][tot], maxi[lhs][i]-mini[rhs][tot-i]);
85 Min(mini[cur][tot], mini[lhs][i]-maxi[rhs][tot-i]);
86 }
87 }
88 }
89 else
90 {
91 //limited negative
92 for (int tot = 0; tot < maxp; ++tot)
93 {
94 //pick mid to be positive
95 for (int i = 0; i <= tot; ++i)
96 {
97 Max(maxi[cur][tot], maxi[lhs][i]+maxi[rhs][tot-i]);
98 Min(mini[cur][tot], mini[lhs][i]+mini[rhs][tot-i]);
99 }
100 }
101 for (int tot = 1; tot < maxp; ++tot)
102 {
103 //pick mid to be negative
104 for (int i = 0; i <= tot-1; ++i)
105 {
106 Max(maxi[cur][tot], maxi[lhs][i]-mini[rhs][tot-1-i]);
107 Min(mini[cur][tot], mini[lhs][i]-maxi[rhs][tot-1-i]);
108 }
109 }
110 }
111 return ret;
112 }
113 int main()
114 {
115 scanf("%s %d %d", s, &p, &m);
116 isp = p <= m;
117 sz = 1;
118 ptr = 0;
119 assert(dfs(1) == p+m);
120
121 printf("%d\n", maxi[1][min(p, m)]);
122 }
WA：+和-写反。dp时候直接按照最大值枚举就好了，不需要特殊处理。
原文地址：https://www.cnblogs.com/skyette/p/8454753.html
时间： 2024-10-02 02:53:00