A. Cards
Description
给定一个仅由onezr组成的字符串，one-1，zero-0。
问能组成的最大的数值是多少。
Solution
模拟贪心。
B. Multiplication Table
Description
给定一个$n \times n$的矩阵m，其主对角线上元素为0。
求一个a向量使得$a[i] \times a[j] = m[i][j] ,i !=j$。
Solution
$a[i] \times a[j] =m[i][j],a[i] \times a[k] = m[i][k] a[j] \times a[k] = m[j][k] \rightarrow a[i]^2=\frac{m[i][j] \times m[i][k]}{m[j][k]}$
C. Substring Game in the Lesson
Description
Solution
显然r不能增大。
那么只要左边存在一个子串字典序小于s[k]，那么Ann跳到最左一个即可。
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(' ')
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 5e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == '-')
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - '0';
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar('-');
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + '0');
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 template <typename T>
72 void _write(const T &t)
73 {
74 write(t);
75 }
76 template <typename T, typename... Args>
77 void _write(const T &t, Args... args)
78 {
79 write(t), pblank;
80 _write(args...);
81 }
82 template <typename T, typename... Args>
83 inline void write_line(const T &t, const Args &... data)
84 {
85 _write(t, data...);
86 }
87 char s[maxn];
88 int pre[maxn], tmp[maxn];
89 int main(int argc, char const *argv[])
90 {
91 #ifndef ONLINE_JUDGE
92 freopen("in.txt", "r", stdin);
93 // freopen("out.txt", "w", stdout);
94 #endif
95 fastIO;
96 cin >> s + 1;
97 n = strlen(s + 1);
98 tmp[0] = 1e9;
99 for (int i = 1; i <= n; i++)
100 tmp[i] = min(tmp[i - 1], (int)s[i]);
101 for (int i = 1; i <= n; i++)
102 {
103 int f = tmp[i] >= s[i];
104 if (f)
105 puts("Mike");
106 else
107 puts("Ann");
108 }
109 return 0;
110 }
D. Alex and Julian
Description
给定一个数字集合S。
当集合内元素满足$|i-j| in S$时，i向j连一条无向边。
求删除最少的数字使集合内剩余的数构成一个二分图(不存在奇环)
Solution
不是很懂还，再想想。
1 #include <algorithm>
2 #include <cctype>
3 #include <cmath>
4 #include <cstdio>
5 #include <cstdlib>
6 #include <cstring>
7 #include <iostream>
8 #include <map>
9 #include <numeric>
10 #include <queue>
11 #include <set>
12 #include <stack>
13 #if __cplusplus >= 201103L
14 #include <unordered_map>
15 #include <unordered_set>
16 #endif
17 #include <vector>
18 #define lson rt << 1, l, mid
19 #define rson rt << 1 | 1, mid + 1, r
20 #define LONG_LONG_MAX 9223372036854775807LL
21 #define pblank putchar(' ')
22 #define ll LL
23 #define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
24 using namespace std;
25 typedef long long ll;
26 typedef long double ld;
27 typedef unsigned long long ull;
28 typedef pair<int, int> P;
29 int n, m, k;
30 const int maxn = 2e5 + 10;
31 template <class T>
32 inline T read()
33 {
34 int f = 1;
35 T ret = 0;
36 char ch = getchar();
37 while (!isdigit(ch))
38 {
39 if (ch == '-')
40 f = -1;
41 ch = getchar();
42 }
43 while (isdigit(ch))
44 {
45 ret = (ret << 1) + (ret << 3) + ch - '0';
46 ch = getchar();
47 }
48 ret *= f;
49 return ret;
50 }
51 template <class T>
52 inline void write(T n)
53 {
54 if (n < 0)
55 {
56 putchar('-');
57 n = -n;
58 }
59 if (n >= 10)
60 {
61 write(n / 10);
62 }
63 putchar(n % 10 + '0');
64 }
65 template <class T>
66 inline void writeln(const T &n)
67 {
68 write(n);
69 puts("");
70 }
71 template <typename T>
72 void _write(const T &t)
73 {
74 write(t);
75 }
76 template <typename T, typename... Args>
77 void _write(const T &t, Args... args)
78 {
79 write(t), pblank;
80 _write(args...);
81 }
82 template <typename T, typename... Args>
83 inline void write_line(const T &t, const Args &... data)
84 {
85 _write(t, data...);
86 }
87 ll a[maxn];
88 vector<int> g[65];
89 int main(int argc, char const *argv[])
90 {
91 #ifndef ONLINE_JUDGE
92 freopen("in.txt", "r", stdin);
93 // freopen("out.txt", "w", stdout);
94 #endif
95 n = read<int>();
96 for (int i = 1; i <= n; i++)
97 {
98 ll x = read<ll>();
99 k = 0;
100 while (x % 2 == 0)
101 {
102 ++k;
103 x /= 2;
104 }
105 g[k].emplace_back(i);
106 }
107 int maxx = 0, f = -1;
108 for (int i = 0; i < 64; i++)
109 if (g[i].size() > maxx)
110 maxx = g[i].size(), f = i;
111 int res = 0;
112 for (int i = 0; i < 64; i++)
113 if (i != f)
114 res += g[i].size();
115 writeln(res);
116 for (int i = 0; i < 64; i++)
117 if (i != f)
118 for (int j = 0; j < g[i].size(); j++)
119 write(g[i][j]), pblank;
120 return 0;
121 }