//嗷嗷嗷
//dfs, 剪枝

C++ 代码

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 500;//多开一点， 最后一天加30会爆
int day[N];
int d[N];
int money[3];
int n;
int res = 0x3f3f3f3f;
void dfs (int u, int v)//u是要走哪一天了，v是到这一天的钱数（不一定是最小的）
{
    if (v >= d[u]) return;//剪枝
    d[u] = v;//存导这一天的最少钱
    if (u > day[n])//全走完辣，开始判断
    {
        res = min (v, res);
        return;
    }
    int t = lower_bound (day + 1, day + 1 + n, u) - day;//二分找到该走哪一天了
    dfs (day[t] + 1, v + money[0]);
    dfs (day[t] + 7, v + money[1]);
    dfs (day[t] + 30, v + money[2]);
}
int main ()
{
    memset (d, 0x3f, sizeof d);
    cin >> n;
    for (int i = 1; i <= n; i ++) cin >> day[i];
    day[n + 1] = 400;
    for (int i = 0; i < 3; i ++) cin >> money[i];
    dfs (day[0], 0);
    cout << res;
    return 0;
}

#include <iostream>
#include <algorithm>
#include <cstring>
using namespace std;
const int N = 500;
int day[N];
int d[N];
int money[3];
int n;
int res = 0x3f3f3f3f;
void dfs (int u, int v)
{
    if (v >= d[u]) return;
    d[u] = v;
    if (u > day[n])
    {
        res = min (v, res);
        return;
    }
    int t = lower_bound (day + 1, day + 1 + n, u) - day;
    dfs (day[t] + 1, v + money[0]);
    dfs (day[t] + 7, v + money[1]);
    dfs (day[t] + 30, v + money[2]);
}
int main ()
{
    memset (d, 0x3f, sizeof d);
    cin >> n;
    for (int i = 1; i <= n; i ++) cin >> day[i];
    day[n + 1] = 400;
    for (int i = 0; i < 3; i ++) cin >> money[i];
    dfs (day[0], 0);
    cout << res;
    return 0;
}

#include <iostream>
#include <cmath>
#include <algorithm>
using namespace std;
const int N = 14;
int f[N];
bool st[N];
int n;
int res = 0;
bool check (int x)
{
    int t = sqrt (x);
    if (t * t == x) return true;
    else return false;
}
void dfs (int u, int v)
{
    if (u == n)
    {
        res ++;
        return;
    }
    for (int i = 1; i <= n; i ++)
    {
        if (i > 1 && !st[i] && !st[i - 1] && f[i] == f[i - 1]) continue;
        //为了避免重复，相同的数下标小的一定在下标大的前面（排序过了）
        //即如果前面的数跟我一样，但是他没被使用，那么我也不能使用
        if (!st[i])
        {
            if (check (f[i] + v))
            {
                st[i] = 1;
                dfs (u + 1, f[i]);
                st[i] = 0;
            }
        }
    }
}
int main ()
{
    cin >> n;
    for (int i = 1; i <= n; i ++) cin >> f[i];
    sort (f + 1, f + 1 + n);
    for (int i = 1; i <= n; i ++)  
    {
        if (i > 1 && f[i] == f[i - 1]) continue;
        st[i] = 1;
        dfs (1, f[i]);
        st[i] = 0;
    }
    cout << res;
    return 0;
}

#include <iostream>
#include <cstring>
using namespace std;
const int N = 210;
bool st[N];
int f[N], d[N];
int n;
int A;
struct point
{
    int a, b, c;
}op[N];
int main ()
{
    cin >> n;
    for (int i = 1; i <= n; i ++)
    {
        scanf ("%d%d", &f[i], &d[i]);
        f[i] *= 2;
        if (!d[i]) A = i;
    }
    for (int t = 0; t < 200; t ++)
    {
        memset (op, 0, sizeof op);
        for (int i = 1; i <= n; i ++)
        {
            if (!st[i]) f[i] += d[i];
            if (f[i] < 2 || f[i] > 198)
            {
                if (i == A)
                {
                    cout << t / 2 + 1;
                    return 0;
                }
                st[i] = 1;
            }
            else
            {
                if (!op[f[i]].a) op[f[i]].a = i;
                else if (!op[f[i]].b) op[f[i]].b = i;
                else op[f[i]].c = i;
            }
        }
        for (int i = 2; i <= 198; i ++)
        {
            if (op[i].a && op[i].b && op[i].c)
            {
                if (!d[op[i].a]) swap (d[op[i].b], d[op[i].c]);
                else if (!d[op[i].b]) swap (d[op[i].a], d[op[i].c]);
                else swap (d[op[i].a], d[op[i].b]);
            }
            else if (op[i].a && op[i].b) swap (d[op[i].a], d[op[i].b]);
        }
    }
    printf ("Cannot fall!");
    return 0;
}

#include <iostream>

using namespace std;

int n;
bool check (int x)
{
    if (x < 2) return false;
    if (x == 2) return true;
    for (int i = 2; i <= x / i; i ++)
        if (x % i == 0) return false;
    return true;
}
int solve (int x)
{
    for (int i = x + 1; 1; i ++)
        if (check (i))
        {
            if (check (i - 6) || check (i + 6)) return i;
        }
}
int main ()
{
    cin >> n;
    if (check (n))
    {
        if (check (n - 6)) cout << "Yes" << endl << n - 6;
        else if (check (n + 6)) cout << "Yes" << endl << n + 6;
        else
        {
            cout << "No" << endl;
            cout << solve (n);
        }
    }
    else
    {
        puts ("No");
        cout << solve (n);
    }
    return 0;
}

#include <iostream>

using namespace std;
const int N = 104708;
int res[N];
bool st[N];
int T;
int cnt = 1;
void check ()
{
    for (int i = 2; i <= N; i ++)
    {
        if (!st[i]) res[cnt ++] = i;
        for (int j = 1; res[j] <= N / i; j ++)
        {
            st[res[j] * i] = true;
            if (i % res[j] == 0) break;
        }
    }
}
int main ()
{
    check ();
    //cout << cnt << endl;
    int x;
    while (cin >> x) printf ("%d\n", res[x]);
    return 0;
}

//闰年：（是4的倍数且不是100的倍数）或 是400的倍数
#include <iostream>

using namespace std;
int a, b;
int check (int x)
{
    int res = 0;
    int year = x / 10000;
    for (int i = 1; i < year; i ++)
    {
        if ((i % 4 == 0 && i % 100 != 0) || i % 400 == 0) res += 366;
        else res += 365;
    }
    int cnt = year * 10000;
    int month = (x - cnt) / 100;
    for (int i = 1; i < month; i ++)
        if (i == 2)
        {
            if ((year % 4 == 0 && year % 100 != 0) || year % 400 == 0) res += 29;
            else res += 28;
        }
        else if (i == 4 || i == 6 || i == 9 || i == 11) res += 30;
        else res += 31;
    res += x - cnt - month * 100;
    //cout << year << ' ' << month << ' ' << x - cnt - month * 100<< ' ' << res << endl;
    return res;
}
int main ()
{
    while (cin >> a >> b)
    {
        if (a > b) swap (a, b);
        printf ("%d\n", check (b) - check (a) + 1);
    }
    return 0;
}

#include <iostream>

using namespace std;
int f[6][6];
bool st[6][6];
int dx[4] = {0, -1, 0, 1}, dy[4] = {1, 0, -1, 0};
int n, m, a, b;
int res = 0x3f3f3f3f;
void dfs (int x, int y, int v, int w)
{
    if (x == a && y == b)
    {
        res = min (res, v);
        return;
    }
    if (v >= res) return;
    for (int i = 0; i < 4; i ++)
    {
        int da = x + dx[i], db = y + dy[i];
        if (da >= 0 && db >= 0 && da < 6 && db < 6 && !st[da][db])
        {
            st[da][db] = true;
            dfs (da, db, v + w * f[da][db], (w * f[da][db]) % 4 + 1);
            st[da][db] = false;
        }
    }
}
int main ()
{
    for (int i = 0; i < 6; i ++)    
        for (int j = 0; j < 6; j ++)
            scanf ("%d", &f[i][j]);
    cin >> n >> m >> a >> b;
    dfs (n, m, 0, 1);
    cout << res;
    return 0;
}

#include <iostream>

using namespace std;
const int N = 510;
int f[N][N];
int n;
int main ()
{
    cin >> n;
    for (int i = 1; i <= n; i ++)
        for (int j = 1; j <= n; j ++)
            scanf ("%d", &f[i][j]);
    int res = 0;
    for (int k = n; k >= 2; k --)
        for (int i = 1; i <= n; i ++)
            for (int j = i + 1; j <= n; j ++)
            {
                f[i][j] = f[j][i] = min (f[i][j], f[i][k] + f[k][j]);
                if (i >= k && j >= k) res += f[i][j] * 2;
            }
    cout << res;

    return 0;
}

#include <iostream>
#include <cstring>
#include <unordered_set>
using namespace std;
typedef unsigned long long ULL;
const int N = 1e4 + 10, M = 131;
char a[N], b[N];
unordered_set<ULL> s;
int n, m;
ULL s1[N], s2[N], p[N];
bool check (int u)
{
    s.clear ();
    for (int i = 1; i + u - 1 <= n; i ++)
    {
        int op = i;
        int j;
        for (j = i; j < i + u; j ++)
            if (a[j] >= '0' && a[j] <= '9')
            {
                i = j;
                break;
            }
        if (j == op + u) s.insert (s1[op + u - 1] - s1[op - 1] * p[u]);
    }
    for (int i = 1; i + u - 1 <= m; i ++)
    {
        int op = i;
        int j;
        for (j = i; j < i + u; j ++)
            if (b[j] >= '0' && b[j] <= '9')
            {
                i = j;
                break;
            }
        if (j == op + u && s.count (s2[op + u - 1] - s2[op - 1] * p[u])) return true;
    }
    return false;
}
int main ()
{
    cin >> a + 1 >> b + 1;
    n = strlen (a + 1), m = strlen (b + 1);
    p[0] = 1;
    for (int i = 1; i <= max (n, m); i ++)
    {
        p[i] = p[i - 1] * M;
        if (i <= n) s1[i] = s1[i - 1] * M + a[i];
        if (i <= m) s2[i] = s2[i - 1] * M + b[i];
    }
    int l = 0, r = min (n, m);
    while (l < r)
    {
        int mid = l + r + 1 >> 1;
        if (check (mid)) l = mid;
        else r = mid - 1;
    }
    cout << l;
    return 0;
}