2018HBCPC部分题解

Mex Query

#include <bits/stdc++.h>
using namespace std;
// 传送门：http://newoj.acmclub.cn/problems/2011
int T;
int n;
int main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> T;
    while (T--)
    {
    
        set<int> s;
        cin >> n;
        for (int i = 1; i <= n; i++)
        {
    
            int remp;
            cin >> remp;
            s.insert(remp);
        }
        set<int>::iterator it;
        int cont = 0;
        for (it = s.begin(); it != s.end(); it++)
        {
    
            if (*it != cont)
            {
    
                cout << cont << endl;
                break;
            }
            cont++;
        }
    }
    return 0;
}   

icebound的商店

#include <bits/stdc++.h>
using namespace std;
// http://newoj.acmclub.cn/problems/2012
// 完全背包
#define mod 1000000009
int bag[15];
int ans[3010];
int T;
int main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    bag[1] = 1;
    bag[2] = 2;
    for (int i = 3; i <= 15; i++)
        bag[i] = bag[i - 1] + bag[i - 2];
    ans[0] = 1;
    for (int i = 1; i <= 15; i++)
    {
    
        for (int j = bag[i]; j <= 3010; j++)
        {
    
            ans[j] += ans[j - bag[i]];
            ans[j] %= mod;
        }
    }
    cin >> T;
    for (int i = 1; i <= T; i++)
    {
    
        int tar;
        cin >> tar;
        cout << ans[tar] << endl;
    }
    return 0;
}

Nim Game

#include <bits/stdc++.h>
using namespace std;
// 博弈论：Nim
// http://newoj.acmclub.cn/problems/2013
const int MOD = 1e9 + 7;
int T;
int n, m;
int a[1000100];
int sum[1000100];
int f[1000100];
int main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> T;
    while (T--)
    {
    
        cin >> n >> m;
        memset(sum, 0, sizeof(sum));
        for (int i = 1; i <= n; i++)
        {
    
            cin >> a[i];
            sum[i] = sum[i - 1] ^ a[i];
        }
        for (int i = 1; i <= m; i++)
        {
    
            int l, r;
            cin >> l >> r;
            int result = sum[r] ^ sum[l - 1]; //第i次结果
            if (result)
                f[i] = 1;
            else
                f[i] = 0;
        }
        long long ans = 0;
        for (int i = 1; i <= m; i++)
        {
    
            ans = ans << 1;
            ans += f[i];
            ans = ans % MOD;
        }
        cout << ans << endl;
    }
    return 0;
}

神殿

#include <bits/stdc++.h>
using namespace std;
// http://newoj.acmclub.cn/problems/2016
long long l, r;
int main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> l >> r;
    while (l <= r)
    {
    
        if ((l | l + 1) > r) //(l | l + 1)为的是将l的最低一位0尝试改成1
            break;
        l = (l | l + 1);
    }
    cout << l << endl;
    return 0;
}

跑图

TLE解法

#include <bits/stdc++.h>
using namespace std;
// TLE解法。。。
int n, m;
int graph[510][510];
int ans[510][510];
int cnt;
struct node
{
    
    int x, y;
} point[250010];
int main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    memset(ans, 0x3f, sizeof(ans));
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
        for (int j = 1; j <= m; j++)
        {
    
            cin >> graph[i][j];
            if (graph[i][j])
            {
    
                ++cnt;
                point[cnt].x = i;
                point[cnt].y = j;
            }
        }
    for (int k = 1; k <= cnt; k++)
    {
    
        for (int i = 1; i <= n; i++)
        {
    
            for (int j = 1; j <= m; j++)
            {
    
                if (graph[i][j])
                    ans[i][j] = 0;
                else
                    ans[i][j] = min(ans[i][j], abs(point[k].x - i) + abs(point[k].y - j));
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
    
        for (int j = 1; j <= m; j++)
        {
    
            cout << ans[i][j];
            if (j != m)
                cout << " ";
        }
        cout << endl;
    }
    return 0;
}

正解

#include <bits/stdc++.h>
using namespace std;
// http://newoj.acmclub.cn/problems/2018
// 正解：BFS
int n, m;
int graph[510][510];
bool visit[510][510];
int ans[510][510];
int dirx[] = {
    0, 0, -1, 1};
int diry[] = {
    1, -1, 0, 0};
struct node
{
    
    int x, y;
};
queue<node> q;
int main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    cin >> n >> m;
    for (int i = 1; i <= n; i++)
    {
    
        for (int j = 1; j <= m; j++)
        {
    
            cin >> graph[i][j];
            if (graph[i][j])
            {
    
                node remp;
                remp.x = i, remp.y = j;
                visit[i][j] = true;
                q.push(remp);
            }
        }
    }
    while (q.empty() == false)
    {
    
        node remp = q.front();
        int rempx = remp.x, rempy = remp.y;
        q.pop();
        for (int i = 0; i < 4; i++)
        {
    
            int nextx = rempx + dirx[i];
            int nexty = rempy + diry[i];
            if (visit[nextx][nexty] == false && (nextx <= n && nextx >= 1 && nexty <= m && nexty >= 1))
            {
    
                ans[nextx][nexty] = ans[rempx][rempy] + 1;
                visit[nextx][nexty] = true; //易错点！！一定要第一次更新完就马上标记！因为此时不标记，后面可能被二次标记，这时候就不是最近的了
                node next;
                next.x = nextx, next.y = nexty;
                q.push(next);
            }
        }
    }
    for (int i = 1; i <= n; i++)
    {
    
        for (int j = 1; j <= m; j++)
        {
    
            cout << ans[i][j];
            if (j != m)
                cout << " ";
        }
        cout << endl;
    }
    return 0;
}

520

#include <bits/stdc++.h>
using namespace std;
// 快速幂
#define MOD 20180520
long long fast_power(long long a, long long b, long long c)
{
    
    long long ans = 1;
    a = a % c;
    while (b)
    {
    
        if (b % 2)
            ans = (ans * a) % c;
        b /= 2;
        a = (a * a) % c;
    }
    return ans;
}
int main()
{
    
    ios::sync_with_stdio(false);
    cin.tie(0), cout.tie(0);
    long long n;
    cin >> n;
    cout << fast_power(2, n, MOD);
    return 0;
}

icebound的账单