1. 这真的是签到题

print("\"ACM welcomes you\\n\"")

2. 4和数

#include <bits/stdc++.h>

using namespace std;

const int maxn = 1e6 + 10;
int f[maxn + 1];
int l, r;

bool check(int num)
{
    int sum = 0;
    while (num)
    {
        sum += num % 10;
        num /= 10;
    }
    if (sum % 4 == 0)
        return true;
    else
        return false;
}

int main()
{
    scanf("%d%d", &l, &r);
    for (int i = 4; i <= maxn; i += 4)
    {
        // 4和数必是4的倍数，缩小范围
        if (check(i))
        {
            // 如果是4和数，标记为1
            f[i] = 1;
        }
    }
    for (int i = 4; i <= maxn; i++)
    {
        // 求前缀和
        f[i] += f[i - 1];
    }
    printf("%d\n", f[r] - f[l - 1]);
    return 0;
}

3. 小D买文具

#include <bits/stdc++.h>
#define maxn 50000

using namespace std;

int n, a, b, c, ans[3];
int m = -1;

int min(int a, int b, int c)
{
    int t;
    if (b > a)
    {
        t = a;
        a = b;
        b = t;
    }
    if (c > b)
    {
        t = b;
        b = c;
        c = t;
    }
    if (b > a)
    {
        t = a;
        a = b;
        b = t;
    }
    return c;
}

int main()
{
    scanf("%d", &n);
    for (int i = 0; i * 5 <= n; i++)
    {
        for (int j = 0; j * 3 <= n - i * 5; j++)
        {
            // 枚举圆规和签字笔，剩余的钱全买稿纸
            if ((n - i * 5 - j * 3) % 2 == 0)
            {
                int k = (n - i * 5 - j * 3) / 2;
                // 如果钱刚好花完
                if (m < min(i, j, k))
                {
                    // 成套的数量尽可能大
                    a = i;
                    b = j;
                    c = k;
                    m = min(i, j, k);
                }
                else if (m == min(i, j, k))
                {
                    if (i + j + k > a + b + c)
                    {
                        // 在满足以上条件情况下，物品的总数尽可能大，即a+b+c尽可能大。
                        a = i;
                        b = j;
                        c = k;
                    }
                }
            }
        }
    }
    printf("%d %d %d", a, b, c);
    return 0;
}

4. 大阳分大洋

#include <bits/stdc++.h>
#define int long long

using namespace std;

const int maxn = 1e6;
int n, x, ans;
int f[maxn]; // 前缀和数组

int get(int l, int r)
{
    // 获取l到r之间有多少大洋
    return f[r] - f[l - 1];
}

signed main()
{
    scanf("%lld", &n);
    for (int i = 1; i <= n; i++)
    {
        scanf("%lld", &x);
        f[i] = f[i - 1] + x; // 求前缀和
    }
    for (int i = 1; i < n; i++)
    {
        // i<n，如果i=n，i+1就大于n了
        if (get(1, i) == get(i + 1, n))
        {
            ans++;
        }
    }
    printf("%d\n", ans);
}

5. 汉诺塔

#include <bits/stdc++.h>
#define mod 1000000007

using namespace std;

typedef long long ll;

int quickpow(ll a, ll b, ll p = mod)
{
    int ans = 1;
    while (b)
    {
        if (b & 1)
            ans = ans * a % p;
        a = a * a % p;
        b >>= 1;
    }
    return ans;
}

int main()
{
    int n;
    scanf("%d", &n);
    ll ans = quickpow(2, n) - 1;
    printf("%d\n", ans);
    return 0;
}

6. 突击考试

#include <bits/stdc++.h>

using namespace std;
const int maxn = 1e5;
int N, K, L = -1;
int A[maxn + 1], B[maxn + 1];

int main()
{
    cin >> N;
    for (int i = 1; i <= N; i++)
    {
        cin >> A[i] >> B[i];
    }

    for (int i = 1; i <= 5; i++)
    {
        int sum = 0;
        for (int j = 1; j <= N; j++)
        {
            if (i == A[j] || i == B[j])
            {
                sum++;
            }
            else
            {
                if (L < sum)
                {
                    L = sum;
                    K = i;
                }
                sum = 0;
            }
            if (L < sum)
            {
                L = sum;
                K = i;
            }
        }
    }
    cout << L << " " << K << endl;
    return 0;
}

7. 小华的立方数

#include <bits/stdc++.h>

using namespace std;
typedef long long ll;

ll getAns(ll k)
{
    ll l = -1;
    ll r = 1e6 + 1;
    while (l < r)
    {
        ll mid = (l + r + 1) / 2;
        if (mid * mid * mid <= k)
        {
            l = mid;
        }
        else if (mid * mid * mid > k)
        {
            r = mid - 1;
        }
    }
    return l;
}

int main()
{
    ll n;
    cin >> n;
    while (n--)
    {
        ll L, R;
        scanf("%lld%lld", &L, &R);
        ll a = getAns(L - 1);
        ll b = getAns(R);
        ll ans = b - a;
        printf("%lld\n", ans);
    }
    return 0;
}

8. 小华的幸福

!> 这道题我只能过70%

#include <bits/stdc++.h>
#define mod 1000000007
#define int long long

typedef long long ll;
using namespace std;

const int maxn = 1e6;
int n, k;
int a[maxn];

int quickpow(int a, int b)
{
    int ans = 1;
    while (b)
    {
        if (b & 1)
        {
            ans = a * ans % mod;
        }
        a = a * a % mod;
        b >>= 1;
    }
    return ans;
}

signed main()
{
    scanf("%lld%lld", &n, &k);
    int fm = ((n - 1) * n) / 2;
    int fz = 0;
    for (int i = 1; i <= n; i++)
    {
        scanf("%lld", &a[i]);
    }
    for (int i = 1; i <= n; i++)
    {
        for (int j = i + 1; j <= n; j++)
        {
            if (a[i] + a[j] == k)
            {
                fz++;
            }
        }
    }
    int ans = quickpow(fm, mod - 2) * fz % mod;
    printf("%lld\n", ans);
    return 0;
}

9. 元素周期表

!> 抄的题解，我读不懂题

#include <iostream>
#include <algorithm>
using namespace std;
const int N = 1e3 + 10;
char s[1010][10] =
    {"1s", "2s", "2p", "3s", "3p", "4s", "3d", "4p", "5s", "4d", "5p", "6s", "4f", "5d", "6p", "7s", "5f", "6d", "7p"};
// s数组优先处理出排列顺序
struct Node
{
    char s[10];
    int cnt;
} a[N];
// 结构体存放每层元素和所具有的原子个数
char st[5] = {'s', 'p', 'd', 'f'}; // 用于对spdf的排序处理
bool cmp(Node x, Node y)
{
    if (x.s[0] != y.s[0])
        return x.s[0] < y.s[0];
    return x.s[1] < y.s[1];
} // 排序规则函数
int main()
{
    int t;
    cin >> t;
    while (t--)
    {
        int n, idx = 0;
        cin >> n;
        for (int i = 0; i < 19; i++)
        {
            a[i].s[0] = s[i][0];
            a[i].s[1] = s[i][1];
            a[i].cnt = 0;
        } // 因为有多组测试用例，初始化结构体
        while (n)
        { // 模拟电子排列顺序
            if (a[idx].s[1] == 's')
            {
                if (n >= 2)
                {
                    a[idx].cnt = 2;
                    n -= 2;
                }
                else
                {
                    a[idx].cnt = n;
                    n = 0;
                }
            }
            else if (a[idx].s[1] == 'p')
            {
                if (n >= 6)
                {
                    a[idx].cnt = 6;
                    n -= 6;
                }
                else
                {
                    a[idx].cnt = n;
                    n = 0;
                }
            }
            else if (a[idx].s[1] == 'd')
            {
                if (n >= 10)
                {
                    a[idx].cnt = 10;
                    n -= 10;
                }
                else
                {
                    a[idx].cnt = n;
                    n = 0;
                }
            }
            else if (a[idx].s[1] == 'f')
            {
                if (n >= 14)
                {
                    a[idx].cnt += 14;
                    n -= 14;
                }
                else
                {
                    a[idx].cnt = n;
                    n = 0;
                }
            }
            idx++; // 逐层填充电子，充当指针作用
        }
        sort(a, a + idx, cmp); // 排序
        for (int i = 0; i < idx; i++)
        {
            if (a[i].cnt == 0)
                continue;
            if (i != 0)
                cout << " ";
            cout << a[i].s[0] << a[i].s[1] << a[i].cnt;
            // 处理输出和格式问题
        }
        cout << endl;
    }
    return 0;
}

10. 剑刃风暴

#include <iostream>
#include <algorithm>
using namespace std;
const int maxn = 1e6 + 5;
struct node
{
    int h; // 血量
    int k; // 亡语数量
} a[maxn];
bool cmp(node x, node y) // 以血量从小到大排序
{
    return x.h < y.h;
}

int idx;

int main()
{
    int n, x;
    cin >> n >> x;
    for (int i = 1; i <= n; i++)
    {
        int A, k;
        cin >> A >> k;
        a[++idx] = {A, k};
        for (int j = 1; j <= k; j++)
        {
            // 因为亡语随从需要等A死后才会收到伤害
            // 所以将亡语所从的血量设置为A+B
            int B;
            cin >> B;
            a[++idx] = {A + B, 0};
        }
    }
    // 按血量排序
    sort(a + 1, a + 1 + idx, cmp);
    // 答案初始化为n
    int ans = n;
    // 开始遍历扣血
    for (int i = 1; i <= idx; i++)
    {
        if (a[i].h != a[i - 1].h) // 如果i和i-1的血量不同，则更新答案
        {
            // 如果剑刃风暴用完了，则不更新答案
            if (x == 0)
                break;
            x--;
        }
        // 每次更新答案 + 亡语随从 - 1（死亡随从）
        ans = ans + a[i].k - 1;
    }
    // 输出答案
    cout << ans;
}