AtCoder Grand Contest 011 题解

A - Airport Bus

贪心，能取就取。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
    char ch;
    int x, a;
    while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
    if (ch == '-') ch = getchar(), a = -1;
    else a = 1;
    x = ch - '0';
    while (ch = getchar(), ch >= '0' && ch <= '9')
        x = (x << 1) + (x << 3) + ch - '0';
    return a * x;
}
 
const int N = 1e5 + 10;
 
int n, c, k, t[N], ans;
 
int main()
{
    giii(n, c, k);
    for (int i = 1; i <= n; ++i) gi(t[i]);
    sort(t + 1, t + n + 1);
    for (int i = 1; i <= n; ++i)
    {
        int j;
        if (t[n] > t[i] + k) j = upper_bound(t + 1, t + n + 1, t[i] + k) - t - 1;
        else j = n;
        j = min(i + c - 1, j);
        i = j;
        ++ans;
    }
    printf("%d
", ans);
}

B - Colorful Creatures

二分答案，直接判断即可。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
    char ch;
    int x, a;
    while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
    if (ch == '-') ch = getchar(), a = -1;
    else a = 1;
    x = ch - '0';
    while (ch = getchar(), ch >= '0' && ch <= '9')
        x = (x << 1) + (x << 3) + ch - '0';
    return a * x;
}
 
const int N = 1e5 + 10;
 
int n;
 
int a[N];
 
bool check(int x)
{
    int64 s = a[x];
    for (int j = 1; j <= n; ++j)
        if (x != j)
        {
            if (s * 2 >= a[j]) s += a[j];
            else return 0;
        }
    return 1;
}

C - Squared Graph

把原图中孤立点、二分图和其它连通块分开考虑一下即可。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
    char ch;
    int x, a;
    while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
    if (ch == '-') ch = getchar(), a = -1;
    else a = 1;
    x = ch - '0';
    while (ch = getchar(), ch >= '0' && ch <= '9')
        x = (x << 1) + (x << 3) + ch - '0';
    return a * x;
}
 
const int N = 1e5 + 10, M = 1e5 + 10;
 
int n, m;
 
bool flag;
 
VI edge[N];
 
int co[N];
 
int O, P, Q;
 
void dfs(int i, int x)
{
    co[i] = x;
    for (auto j : edge[i])
    {
        if (co[j])
        {
            if (co[j] != (co[i] ^ 1)) flag = 0;
        }
        else
        {
            dfs(j, x ^ 1);
        }
    }
}
 
int main()
{
    gii(n, m);
    for (int i = 1; i <= m; ++i)
    {
        int u, v;
        gii(u, v);
        edge[u].pb(v);
        edge[v].pb(u);
    }
    for (int i = 1; i <= n; ++i)
    {
        if (!co[i])
        {
            if (!SZ(edge[i]))
            {
                ++O;
                continue;
            }
            flag = 1;
            dfs(i, 2);
            if (flag) ++Q;
            else ++P;
        }
    }
    long long ans = 2LL * O * n - 1LL * O * O + 1LL * P * P + 2LL * P * Q + 2LL * Q * Q;
    printf("%lld
", ans);
    return 0;
}

D - Half Reflector

我们发现2*n次之后肯定就是一个循环了。前面的暴力即可。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
    char ch;
    int x, a;
    while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
    if (ch == '-') ch = getchar(), a = -1;
    else a = 1;
    x = ch - '0';
    while (ch = getchar(), ch >= '0' && ch <= '9')
        x = (x << 1) + (x << 3) + ch - '0';
    return a * x;
}
 
const int N = 2e5 + 10;
 
char str[N];
 
int n, k;
 
int pos(int x, int l)
{
    int p = x + l % n;
    if (p > n) p -= n;
    return p;
}
 
int main()
{
    gii(n, k);
    scanf("%s", str + 1);
    for (int i = 1; i <= n; ++i) str[i] -= 'A';
    if (k <= 2 * n + 10)
    {
        int l = 0, t = 0;
        for (int i = 1; i <= k; ++i)
        {
            if (!(str[pos(1, l)] ^ t))
            {
                str[pos(1, l)] ^= 1;
            }
            else
            {
                ++l, t ^= 1; str[pos(n, l)] = t;
            }
        }
        for (int i = 1; i <= n; ++i) putchar((str[pos(i, l)] ^ t) + 'A');
        puts("");
        return 0;
    }
    else
    {
        int l = 0, t = 0;
        for (int i = 1; i <= 2 * n + 10; ++i)
        {
            if (!(str[pos(1, l)] ^ t))
            {
                str[pos(1, l)] ^= 1;
            }
            else
            {
                ++l, t ^= 1; str[pos(n, l)] = t;
            }
        }
        k -= 2 * n + 10;
        if (n & 1)
        {
            str[pos(1, l)] ^= (k & 1);
            for (int i = 1; i <= n; ++i) putchar((str[pos(i, l)] ^ t) + 'A');
            puts("");
            return 0;
        }
        else
        {
            for (int i = 1; i <= n; ++i) putchar((str[pos(i, l)] ^ t) + 'A');
            puts("");
            return 0;
        }
    } 
}

E - Increasing Numbers

我们发现一个上升的数肯定可以拆成很多个不同位数的111111……之和。就是(10^x-1)/9，我们暴力加一下，直接判断即可。

//waz
#include <bits/stdc++.h>
 
using namespace std;
 
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define ALL(x) (x).begin(), (x).end()
#define SZ(x) ((int)((x).size()))
 
typedef pair<int, int> PII;
typedef vector<int> VI;
typedef long long int64;
typedef unsigned int uint;
typedef unsigned long long uint64;
 
#define gi(x) ((x) = F())
#define gii(x, y) (gi(x), gi(y))
#define giii(x, y, z) (gii(x, y), gi(z))
 
int F()
{
    char ch;
    int x, a;
    while (ch = getchar(), (ch < '0' || ch > '9') && ch != '-');
    if (ch == '-') ch = getchar(), a = -1;
    else a = 1;
    x = ch - '0';
    while (ch = getchar(), ch >= '0' && ch <= '9')
        x = (x << 1) + (x << 3) + ch - '0';
    return a * x;
}
 
int n;
 
char s[500010];
 
int x[500010];
 
int sum;
 
int main()
{
    scanf("%s", s + 1);
    n = strlen(s + 1);
    reverse(s + 1, s + n + 1);
    for (int i = 1; i <= n; ++i) x[i] = (s[i] - 48) * 9;
    for (int i = 1; i <= n; ++i) x[i + 1] += x[i] / 10, x[i] %= 10;
    if (x[n + 1]) ++n;
    for (int i = 1; i <= n; ++i) sum += x[i];
    for (int k = 1; k <= 2 * n; ++k)
    {
        int v = 9, j = 1;
        while (v)
        {
            sum -= x[j];
            x[j] += v;
            v = x[j] / 10;
            x[j] %= 10;
            sum += x[j];
            if (j > n) ++n;
            ++j;
        }
        if (sum <= 9 * k) 
        {
            printf("%d
", k);
            return 0;
        }
    }
    return 0;
}

F - Train Service Planning

题目都不是很懂（太菜了qwq）。。。先咕着吧。