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 ", 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 ", 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 ", 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 ", 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 ", 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。