codeforces #586 ABC~D

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$n imes n$的矩阵m，其主对角线上元素为0。

求一个a向量使得$a[i] imes a[j] = m[i][j] ,i !=j$。

Solution

$a[i] imes a[j] =m[i][j],a[i] imes a[k] = m[i][k] a[j] imes a[k] = m[j][k] ightarrow a[i]^2=frac{m[i][j] imes m[i][k]}{m[j][k]}$

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 }

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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 }

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