牛客国庆集训派对Day5 Solution

A    璀璨光滑

留坑。

 

B    电音之王

蒙特马利大数乘模运算

#include <bits/stdc++.h>
using namespace std;

typedef long long ll;
typedef unsigned long long u64;
typedef __int128_t i128;
typedef __uint128_t u128;

struct Mod64 {
    Mod64() :n_(0) {}
    Mod64(u64 n) :n_(init(n)) {}
    static u64 init(u64 w) { return reduce(u128(w) * r2); }
    static void set_mod(u64 m) {
        mod = m; assert(mod & 1);
        inv = m; for (int i = 0; i < 5; ++i) inv *= 2 - inv * m;
        r2 = -u128(m) % m;
    }
    static u64 reduce(u128 x) {
        u64 y = u64(x >> 64) - u64((u128(u64(x)*inv)*mod) >> 64);
        return ll(y)<0 ? y + mod : y;
    }
    Mod64& operator += (Mod64 rhs) { n_ += rhs.n_ - mod; if (ll(n_)<0) n_ += mod; return *this; }
    Mod64 operator + (Mod64 rhs) const { return Mod64(*this) += rhs; }
    Mod64& operator -= (Mod64 rhs) { n_ -= rhs.n_; if (ll(n_)<0) n_ += mod; return *this; }
    Mod64 operator - (Mod64 rhs) const { return Mod64(*this) -= rhs; }
    Mod64& operator *= (Mod64 rhs) { n_ = reduce(u128(n_)*rhs.n_); return *this; }
    Mod64 operator * (Mod64 rhs) const { return Mod64(*this) *= rhs; }
    u64 get() const { return reduce(n_); }
    static u64 mod, inv, r2;
    u64 n_;
};

u64 Mod64::mod, Mod64::inv, Mod64::r2;


int t, k;
u64 A0, A1, M0, M1, C, M;

void Run()
{
    scanf("%d", &t);
    while (t--)
    {
        scanf("%llu%llu%llu%llu%llu%llu%d", &A0, &A1, &M0, &M1, &C, &M, &k);
        Mod64::set_mod(M);
        Mod64 a0(A0), a1(A1), m0(M0), m1(M1), c(C), ans(1), a2(0);
        ans *= a0; ans *= a1;
        for (int i = 2; i <= k; ++i)
        {
            a2 = a1;
            a1 = m0 * a1 + m1 * a0 + c;
            a0 = a2;
            ans *= a1;
        }
        printf("%llu\n", ans.get());
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif

    Run();
    return 0;
}

 

 

C    萌新拆塔

留坑。

 

D    奇迹暖婊

留坑。

 

E    风花雪月

留坑。

 

F    双倍掉率

留坑。

 

G    贵族用户

枚举VIP等级，注意精度问题 不要用(1-p * 0.01) 用 (100 - p) * 1.0 / 100

#include<bits/stdc++.h>
 
using namespace std;
 
const int maxn = 1e2 + 10;
 
int m, k;
int a[maxn], p[maxn], c[maxn], d[maxn];
 
int main()
{
    int t;
    scanf("%d", &t);
    while(t--)
    {
        scanf("%d %d", &m, &k);
        for(int i = 1; i <= m; ++i) scanf("%d %d", a + i, p + i);
        for(int i = 1; i <= k; ++i) scanf("%d %d", c + i, d + i);
        int ans = 0x3f3f3f3f;
        //for(int i = 1; i <= k; ++i) ans += c[i] * d[i];
        for(int i = 0; i <= m; ++i)
        {
            int res = 0;
            for(int j = 1; j <= k; ++j)
            {
                int tmp = (int)ceil(d[j] * (100.0 - p[i]) / 100.0);
                res += tmp * c[j];
            }
            if(res < a[i]) res = a[i];
            ans = min(ans, res);
        }
        ans = (int)ceil(ans * 0.1);
        printf("%d\n", ans);
    }
    return 0;
}

 

H    我不爱她

留坑。

 

I    人渣本愿

留坑。

 

J    友谊巨轮

用线段树维护每个人对其他人的聊天记录，然后取max上去，但是开1e5棵线段树肯定空间爆炸，考虑动态开点，每次最多开logn个结点

#include <bits/stdc++.h>
using namespace std;

#define N 100010
#define ll long long

int t, n, m, k;
int root[N], res;  

struct SEG
{
    #define M N * 35
    int cnt, lson[M], rson[M];
    struct node
    {
        int tm, pos;
        ll Max;
        bool operator < (const node &r) const
        {
            return Max == r.Max ? tm < r.tm : Max < r.Max;
        }
    }a[M];

    void pushup(int id)
    {
        a[id] = max(a[lson[id]], a[rson[id]]);
    }

    void update(int &root, int l, int r, ll val, int tm, int pos)
    {
        if (!root)
        {
            root = ++cnt;
            lson[root] = rson[root] = 0;
            a[root].Max = a[root].tm = a[root].pos = 0;
        }
        if (l == r)
        {
            a[root].Max += val;
            a[root].tm = max(a[root].tm, tm);
            a[root].pos = pos;
            return;
        }
        int mid = (l + r) >> 1;
        if (pos <= mid) update(lson[root], l, mid, val, tm, pos);
        else update(rson[root], mid + 1, r, val, tm, pos);
        pushup(root);
    }
}segtree;

void Init()
{
    segtree.cnt = 0;
    memset(root, 0, sizeof root);
    res = 0;
}

struct OP
{
    int a, b, c, tm;
    void scan(int tm)
    {
        this->tm = tm;
        scanf("%d%d%d", &a, &b, &c);
    }
}op[N];

void update(int a, int b, int c, int tm)
{
    int bef = segtree.a[root[a]].pos;
    segtree.update(root[a], 1, n, c, tm, b);
    int now = segtree.a[root[a]].pos;
    if (bef == now) return;
    if (bef) 
    {
        if (segtree.a[root[bef]].pos == a) ++res;
        else --res;
    }
    if (now)
    {
        if (segtree.a[root[now]].pos == a) --res;
        else ++res; 
    }

}

void Run()
{
    segtree.a[0].tm = 1000000;  // 防止取max操作时误把空节点pushup上去
    scanf("%d", &t);
    while (t--)
    {
        Init();
        scanf("%d%d%d", &n, &k, &m);
        for (int i = 1; i <= k; ++i)
        {
            op[i].scan(i);
            update(op[i].a, op[i].b, op[i].c, i);
            update(op[i].b, op[i].a, op[i].c, i);
            if (i > m)
            { 
                update(op[i - m].a, op[i - m].b, -op[i - m].c, i - m);
                update(op[i - m].b, op[i - m].a, -op[i - m].c, i - m);
            }
            printf("%d\n", res);
        }
    }
}

int main()
{
    #ifdef LOCAL
        freopen("Test.in", "r", stdin);
    #endif

    Run();
    return 0;
}

 

 

K    最后战役

留坑。

 

L    数论之神

两个规律，在代码中。

#include <bits/stdc++.h>
using namespace std;
 
#define ll long long
 
int T;
ll n, k;
 
int main()
{
    scanf("%d", &T);
    while (T--)
    {
        scanf("%lld%lld", &n, &k);
        ll t = (ll)sqrt(n);
//      cout << t << endl;
        ll num = t * (t + 1) <= n ? 2 * t : 2 * t - 1;
        k = num - k + 1;
        ll K = k <= t ? k : n / (num - k + 1);
        printf("%lld %lld\n", num, K);
    }
    return 0;
}