ACM模版

这个比赛早就知道有,但是因为自己要骑行,结果就没有注册,后来骑行计划延期,但是也忘了注册,赛后重现赛尝试做了 12 道,感觉水题比较多,剩下三个 AC 的人好少啊,感觉我这么菜肯定也是做不出来,所以就先不补了吧……

A-黑白图像直方图

描述

题解

水题,扫描一遍就行了。

代码

#include <iostream>
#include <cstdio>
#include <cstring>

using namespace std;

const int MAXN = 111;

int n, m;
int cnt[MAXN];

int main(int argc, const char * argv[])
{
    while (cin >> n >> m)
    {
        memset(cnt, 0, sizeof(cnt));

        int x;
        for (int i = 0; i < n; i++)
        {
            for (int j = 0; j < m; j++)
            {
                scanf("%d", &x);
                if (x)
                {
                    cnt[j]++;
                }
            }
        }

        printf("%d", cnt[0]);
        for (int j = 1; j < m; j++)
        {
            printf(" %d", cnt[j]);
        }
        putchar(10);
    }

    return 0;
}

B-神无月排位赛

描述

题解

模拟题,注意细节就好了。

代码

#include <iostream>

using namespace std;

int N;

int main(int argc, const char * argv[])
{
    int level, score;
    while (cin >> N)
    {
        level = 3;
        score = 0;

        int x, y;
        for (int i = 0; i < N; i++)
        {
            scanf("%d", &x);
            if (x)
            {
                score += 10;
            }
            else
            {
                score -= 5;
            }
            if (score < 0)
            {
                score = 0;
            }
            if (score >= 100)
            {
                if (i + 2 >= N)
                {
                    continue;
                }
                scanf("%d%d", &x, &y);
                if (x == y)
                {
                    if (x)
                    {
                        level--;
                        score = 0;
                    }
                    else
                    {
                        score = 60;
                    }
                    i += 2;
                    continue;
                }
                if (i + 3 >= N)
                {
                    continue;
                }
                scanf("%d", &x);
                if (x)
                {
                    level--;
                    score = 0;
                }
                else
                {
                    score = 60;
                }
                i += 3;
            }
        }

        if (level < 0)
        {
            puts("S");
        }
        else
        {
            putchar('A' + level);
            putchar(10);
        }
    }

    return 0;
}

C-I Love ces

描述

题解

扫描一遍,计数就行了。

代码

#include <iostream>
#include <cstring>
#include <string>

using namespace std;

const int MAXN = 100;
const int MAGIC = 32;
const int INF = 0x3f3f3f3f;

string s;
int cnt[MAXN];

int main(int argc, const char * argv[])
{
    while (cin >> s)
    {
        memset(cnt, 0, sizeof(cnt));

        int len = (int)s.length();
        for (int i = 0; i < len; i++)
        {
            if (s[i] >= 'a')
            {
                s[i] -= MAGIC;
            }
            cnt[s[i]]++;
        }
        cnt['E'] /= 2;

        int res = INF;
        res = min(res, cnt['I']);
        res = min(res, cnt['L']);
        res = min(res, cnt['O']);
        res = min(res, cnt['V']);
        res = min(res, cnt['E']);
        res = min(res, cnt['C']);
        res = min(res, cnt['S']);

        cout << res << '\n';
    }

    return 0;
}

D-添加好友

描述

题解

简单的组合,输出 2n1 2 n − 1 即可,快速幂。

代码

#include <iostream>

using namespace std;

const int MOD = 1e9 + 7;

typedef long long ll;

ll QPow(ll x, ll n)
{
    ll ret = 1;
    ll tmp = x % MOD;

    while (n)
    {
        if (n & 1)
        {
            ret = (ret * tmp) % MOD;
        }
        tmp = tmp * tmp % MOD;
        n >>= 1;
    }

    return ret;
}

int n;

int main(int argc, const char * argv[])
{
    while (cin >> n)
    {
        cout << QPow(2, n) - 1 << '\n';
    }

    return 0;
}

E-字符串进制转换

描述

题解

数学题,先转化为十进制,然后短除法即可。

代码

#include <iostream>
#include <algorithm>
#include <string>

using namespace std;

typedef long long ll;

const int MAXN = 15;
const int MAGIC = 26;

int m;
ll POW[MAXN] = {
  1};
string A;

void init()
{
    for (int i = 1; i < MAXN; i++)
    {
        POW[i] = POW[i - 1] * MAGIC;
    }
}

int main(int argc, const char * argv[])
{
    init();

    while (cin >> A >> m)
    {
        ll num = 0;
        for (int i = (int)A.length() - 1, j = 0; i >= 0; i--, j++)
        {
            num += (A[i] - 'a') * POW[j];
        }

        if (num == 0)
        {
            cout << "0\n";
            continue;
        }

        string ans;
        while (num)
        {
            ans.push_back(num % m + '0');
            num /= m;
        }
        reverse(ans.begin(), ans.end());

        cout << ans << '\n';
    }

    return 0;
}

F-A序列

描述

题解

说白了,就是左右各一遍 LIS 即可。模版题~~~

代码

#include <cstdio>
#include <algorithm>
#include <iostream>

using namespace std;

/* * 递增(默认) * 递减 * 非递增 * 非递减 (1)>= && < (2)< (3)>= */
const int MAXN = 5e5 + 10;

int a[MAXN], f[MAXN], d[MAXN], d_[MAXN];   // d[i] 用于记录 a[0...i] 以 a[i] 结尾的最大长度

int bsearch(const int *f, int size, const int &a)
{
    int l = 0, r = size - 1;
    while (l <= r)
    {
        int mid = (l + r) / 2;
        if (a > f[mid - 1] && a <= f[mid])  // (1)
        {
            return mid;
        }
        else if (a < f[mid])
        {
            r = mid - 1;
        }
        else
        {
            l = mid + 1;
        }
    }
    return -1;
}

int LIS(const int *a, int *d, const int &n)
{
    int i, j, size = 1;
    f[0] = a[0];
    d[0] = 1;
    for (i = 1; i < n; ++i)
    {
        if (a[i] <= f[0])               // (2)
        {
            j = 0;
        }
        else if (a[i] > f[size - 1])    // (3)
        {
            j = size++;
        }
        else
        {
            j = bsearch(f, size, a[i]);
        }
        f[j] = a[i];
        d[i] = j + 1;
    }
    return size;
}

int main()
{
    int i, n;
    while (scanf("%d", &n) != EOF)
    {
        for (i = 0; i < n; ++i)
        {
            scanf("%d", &a[i]);
        }
        LIS(a, d, n);
        reverse(a, a + n);
        LIS(a, d_, n);

        int res = 1;
        for (int i = 0; i < n; i++)
        {
            if (d[i] <= d_[n - i - 1])
            {
                res = max(res, (d[i] << 1) - 1);
            }
            else
            {
                res = max(res, (d_[n - i - 1] << 1) - 1);
            }
        }

        cout << res << '\n';
    }

    return 0;
}

G-战斗

描述

题解

直接暴力,dfs 出来所有情况,判断是否有合理情况存在即可。

代码

#include <iostream>
#include <cstring>

using namespace std;

const int MAXN = 11;

struct node
{
    int v, a;
} my[MAXN], sy[MAXN];

int n;
int rk[MAXN], vis[MAXN];

bool judge()
{
    int i = 0, j = 0;
    node a = sy[i], b = my[rk[j]];
    while (i < n && j < n)
    {
        int t1 = a.v / b.a;
        int t2 = b.v / a.a;
        int x = min(t1, t2);
        a.v -= b.a * x;
        b.v -= a.a * x;
        while (a.v > 0 && b.v > 0)
        {
            a.v -= b.a;
            b.v -= a.a;
        }
        if (a.v <= 0)
        {
            a = sy[++i];
        }
        if (b.v <= 0)
        {
            b = my[rk[++j]];
        }
    }
    return j == n ? false : true;
}

bool dfs(int pos)
{
    if (pos == n)
    {
        if (judge())
        {
            return true;
        }
        return false;
    }

    for (int i = 0; i < n; i++)
    {
        if (!vis[i])
        {
            rk[pos] = i;
            vis[i] = 1;
            if (dfs(pos + 1))
            {
                return true;
            }
            vis[i] = 0;
        }
    }

    return false;
}

int main()
{
    int T;
    scanf("%d", &T);
    while (T--)
    {
        memset(vis, 0, sizeof(vis));

        scanf("%d", &n);
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d", &sy[i].v, &sy[i].a);
        }
        for (int i = 0; i < n; i++)
        {
            scanf("%d%d", &my[i].v, &my[i].a);
        }

        if (dfs(0))
        {
            printf("YES\n");
        }
        else
        {
            printf("NO\n");
        }
    }

    return 0;
}

H-调和序列

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I-丢史蒂芬妮

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J-膜一下将带给你好运

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