不完全模板如下，长期更新中(已弃):

静态内存指针版动态开点线段树…(这名字有毒)

自行修改 $pushDown$ 和 $update$ 等…

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27

struct Node {
    Node *lc, *rc;
    int l, r;
    long long val;
    int tag;
    inline void clear() { val = tag = 0; }
    inline void cover(const int delta) { tag += delta, val += (long long)delta * (r - l + 1); }
    inline void pushDown() { if (tag) lc->cover(tag), rc->cover(tag), tag = 0; }
    inline void update(const int l, const int r, const int delta) {
        if (l > this->r || r < this->l) return;
        else if (l <= this->l && r >= this->r) cover(delta);
        else pushDown(), lc->update(l, r, delta), rc->update(l, r, delta), val = lc->val + rc->val;
    }
    inline long long query(const int l, const int r) {
        register long long ret = 0;
        if (l > this->r || r < this->l) return 0;
        else if (l <= this->l && r >= this->r) return val;
        else return pushDown(), ret += lc->query(l, r), ret += rc->query(l, r), ret;
    }
} pool[(MAXN + 10) << 4], *root, *tail = pool;
inline void build(Node *&rt, const int l, const int r, int *a) {
    rt = tail++, rt->clear(), rt->l = l, rt->r = r;
    if (l == r) { rt->val = a[l]; return; }
    register int mid = l + r >> 1;
    build(rt->lc, l, mid, a), build(rt->rc, mid + 1, r, a);
    rt->val = rt->lc->val + rt->rc->val;
}

质数,欧拉函数及莫比乌斯函数的快速线性筛选法，时间复杂度 $O(n)$

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16

const int MAXN = 100000;
int phi[MAXN + 10], prime[MAXN + 10], mu[MAXN + 10], tot;
bool vis[MAXN + 10];
inline void init() {
    mu[1] = phi[1] = 1;
    for (register int i = 2; i <= MAXN; i++) {
        if (!vis[i]) prime[++tot] = i, phi[i] = i - 1, mu[i] = -1;
        for (register int j = 1; j <= tot && i * prime[j] <= MAXN; j++) {
            vis[i * prime[j]] = 1;
            if (i % prime[j] == 0) {
                phi[i * prime[j]] = phi[i] * prime[j], mu[i * prime[j]] = 0;
                break;
            } else phi[i * prime[j]] = phi[i] * (prime[j] - 1), mu[i * prime[j]] = -mu[i];
        }
    }
}

Exponentiation of one integer by another often produces very large results. In this problem, we will compute a function based on repeated exponentiation, but output only the last n digits of the result. Doing this efficiently requires careful thought about how to avoid computing the full answer.

Given integers b, n, and i, we define the function f(x) recursively by f(x) = b f(x-1) if x > 0, and f(0)=1. Your job is to efficiently compute the last n decimal digits of f(i).

你被要求设计一个计算器完成以下三项任务：

1,给定y,z,p,计算Y^Z Mod P 的值；

2,给定y,z,p，计算满足xy≡ Z ( mod P )的最小非负整数；

3,给定y,z,p，计算满足Y^x ≡ Z ( mod P)的最小非负整数。

We say that integer x, 0 < x < p, is a primitive root modulo odd prime p if and only if the set { (xi mod p) | 1 <= i <= p-1 } is equal to { 1, …, p-1 }. For example, the consecutive powers of 3 modulo 7 are 3, 2, 6, 4, 5, 1, and thus 3 is a primitive root modulo 7.
Write a program which given any odd prime 3 <= p < 65536 outputs the number of primitive roots modulo p.

Little Y finds there is a very interesting formula in mathematics:

$X^Y$ mod $Z = K$

Given X, Y, Z, we all know how to figure out K fast. However, given X, Z, K, could you figure out Y fast?

Sheng bill有着惊人的心算能力，甚至能用大脑计算出两个巨大的数的GCD（最大公约 数）！因此他经常和别人比赛计算GCD。有一天Sheng bill很嚣张地找到了你，并要求和你比 赛，但是输给Sheng bill岂不是很丢脸！所以你决定写一个程序来教训他。

Elina is reading a book written by Rujia Liu, which introduces a strange way to express non-negative integers. The way is described as following:
Choose k different positive integers a1, a2, \cdots , ak. For some non-negative m, divide it by every ai (1 \leq i \leq k) to find the remainder ri. If a1, a2, \cdots , ak are properly chosen, m can be determined, then the pairs (ai, ri) can be used to express m.

“It is easy to calculate the pairs from m, ” said Elina. “But how can I find m from the pairs?”

Since Elina is new to programming, this problem is too difficult for her. Can you help her?

Some people believe that there are three cycles in a person’s life that start the day he or she is born. These three cycles are the physical, emotional, and intellectual cycles, and they have periods of lengths 23, 28, and 33 days, respectively. There is one peak in each period of a cycle. At the peak of a cycle, a person performs at his or her best in the corresponding field (physical, emotional or mental). For example, if it is the mental curve, thought processes will be sharper and concentration will be easier.

A Compiler Mystery: We are given a C-language style for loop of type
for (variable = A; variable != B; variable += C)

statement;

I.e., a loop which starts by setting variable to value A and while variable is not equal to B, repeats statement followed by increasing the variable by C. We want to know how many times does the statement get executed for particular values of A, B and C, assuming that all arithmetics is calculated in a k-bit unsigned integer type (with values 0 <= x < 2 k) modulo 2 k.

One day Vasya was sitting on a not so interesting Maths lesson and making an origami from a rectangular a mm  ×  b mm sheet of paper (a > b). Usually the first step in making an origami is making a square piece of paper from the rectangular sheet by folding the sheet along the bisector of the right angle, and cutting the excess part.

A lattice point (x, y) in the first quadrant (x and y are integers greater than or equal to 0), other than the origin, is visible from the origin if the line from (0, 0) to (x, y) does not pass through any other lattice point. For example, the point (4, 2) is not visible since the line from the origin passes through (2, 1). The figure below shows the points (x, y) with 0 \leq x, y \leq 5 with lines from the origin to the visible points.

Given a positive integer X, an X-factor chain of length m is a sequence of integers,

$1 = X_0, X_1, X_2, \cdots , X_m = X$

satisfying

$X_i < X_i+1$ and $X_i | X_i+1$ where a | b means a perfectly divides into b.

Now we are interested in the maximum length of X-factor chains and the number of chains of such length.

The branch of mathematics called number theory is about properties of numbers. One of the areas that has captured the interest of number theoreticians for thousands of years is the question of primality. A prime number is a number that is has no proper factors (it is only evenly divisible by 1 and itself). The first prime numbers are 2,3,5,7 but they quickly become less frequent. One of the interesting questions is how dense they are in various ranges. Adjacent primes are two numbers that are both primes, but there are no other prime numbers between the adjacent primes. For example, 2,3 are the only adjacent primes that are also adjacent numbers.
Your program is given 2 numbers: L and U (1<=L< U<=2,147,483,647), and you are to find the two adjacent primes C1 and C2 (L<=C1< C2<=U) that are closest (i.e. C2-C1 is the minimum). If there are other pairs that are the same distance apart, use the first pair. You are also to find the two adjacent primes D1 and D2 (L<=D1< D2<=U) where D1 and D2 are as distant from each other as possible (again choosing the first pair if there is a tie).

Om Nom knows that his girlfriend loves beautiful patterns. That’s why he wants the beads on the necklace to form a regular pattern. A sequence of beads S is regular if it can be represented as S = A + B + A + B + A + … + A + B + A, where A and B are some bead sequences, “ + “ is the concatenation of sequences, there are exactly 2k + 1 summands in this sum, among which there are k + 1 “A” summands and k “B” summands that follow in alternating order. Om Nelly knows that her friend is an eager mathematician, so she doesn’t mind if A or B is an empty sequence.