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$n \times n$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。

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#include <vector>
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define LONG_LONG_MAX 9223372036854775807LL
#define pblank putchar(' ')
#define ll LL
#define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> P;
int n, m, k;
const int maxn = 5e5 + 10;
template <class T>
inline T read()
{
    int f = 1;
    T ret = 0;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        ret = (ret << 1) + (ret << 3) + ch - '0';
        ch = getchar();
    }
    ret *= f;
    return ret;
}
template <class T>
inline void write(T n)
{
    if (n < 0)
    {
        putchar('-');
        n = -n;
    }
    if (n >= 10)
    {
        write(n / 10);
    }
    putchar(n % 10 + '0');
}
template <class T>
inline void writeln(const T &n)
{
    write(n);
    puts("");
}
template <typename T>
void _write(const T &t)
{
    write(t);
}
template <typename T, typename... Args>
void _write(const T &t, Args... args)
{
    write(t), pblank;
    _write(args...);
}
template <typename T, typename... Args>
inline void write_line(const T &t, const Args &... data)
{
    _write(t, data...);
}
char s[maxn];
int pre[maxn], tmp[maxn];
int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif
    fastIO;
    cin >> s + 1;
    n = strlen(s + 1);
    tmp[0] = 1e9;
    for (int i = 1; i <= n; i++)
        tmp[i] = min(tmp[i - 1], (int)s[i]);
    for (int i = 1; i <= n; i++)
    {
        int f = tmp[i] >= s[i];
        if (f)
            puts("Mike");
        else
            puts("Ann");
    }
    return 0;
}

 

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。

 

 

#include <algorithm>
#include <cctype>
#include <cmath>
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <iostream>
#include <map>
#include <numeric>
#include <queue>
#include <set>
#include <stack>
#if __cplusplus >= 201103L
#include <unordered_map>
#include <unordered_set>
#endif
#include <vector>
#define lson rt << 1, l, mid
#define rson rt << 1 | 1, mid + 1, r
#define LONG_LONG_MAX 9223372036854775807LL
#define pblank putchar(' ')
#define ll LL
#define fastIO ios::sync_with_stdio(false), cin.tie(0), cout.tie(0)
using namespace std;
typedef long long ll;
typedef long double ld;
typedef unsigned long long ull;
typedef pair<int, int> P;
int n, m, k;
const int maxn = 2e5 + 10;
template <class T>
inline T read()
{
    int f = 1;
    T ret = 0;
    char ch = getchar();
    while (!isdigit(ch))
    {
        if (ch == '-')
            f = -1;
        ch = getchar();
    }
    while (isdigit(ch))
    {
        ret = (ret << 1) + (ret << 3) + ch - '0';
        ch = getchar();
    }
    ret *= f;
    return ret;
}
template <class T>
inline void write(T n)
{
    if (n < 0)
    {
        putchar('-');
        n = -n;
    }
    if (n >= 10)
    {
        write(n / 10);
    }
    putchar(n % 10 + '0');
}
template <class T>
inline void writeln(const T &n)
{
    write(n);
    puts("");
}
template <typename T>
void _write(const T &t)
{
    write(t);
}
template <typename T, typename... Args>
void _write(const T &t, Args... args)
{
    write(t), pblank;
    _write(args...);
}
template <typename T, typename... Args>
inline void write_line(const T &t, const Args &... data)
{
    _write(t, data...);
}
ll a[maxn];
vector<int> g[65];
int main(int argc, char const *argv[])
{
#ifndef ONLINE_JUDGE
    freopen("in.txt", "r", stdin);
    // freopen("out.txt", "w", stdout);
#endif
    n = read<int>();
    for (int i = 1; i <= n; i++)
    {
        ll x = read<ll>();
        k = 0;
        while (x % 2 == 0)
        {
            ++k;
            x /= 2;
        }
        g[k].emplace_back(i);
    }
    int maxx = 0, f = -1;
    for (int i = 0; i < 64; i++)
        if (g[i].size() > maxx)
            maxx = g[i].size(), f = i;
    int res = 0;
    for (int i = 0; i < 64; i++)
        if (i != f)
            res += g[i].size();
    writeln(res);
    for (int i = 0; i < 64; i++)
        if (i != f)
            for (int j = 0; j < g[i].size(); j++)
                write(g[i][j]), pblank;
    return 0;
}