数学章节：筛素数、分解质因数、

数论知识总结

分类：筛素数（埃氏筛法、线性筛法、二次筛法）、分解质因数、快速幂、约数个数、欧拉函数、同余、矩阵乘法、组合计数、高斯消元、容斥原理、概率与数学期望、博弈论

筛素数方法

埃氏筛法： $l o g l o g (n)$ ，素数所有的倍数直接筛掉，线性筛法： $O (n)$ ，不断素数倍数筛掉，二次筛法间接地把某一段很大的区间素数筛掉（线性筛法时间复杂度较高的情况）

注意的是，1有的时候需要特判，联系图论一起考察

1292. 哥德巴赫猜想

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1e6 + 10;
bool st[N];
int primes[N], cnt;
int n;

void init(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    init(N - 1);
    
    while (cin >> n, n)
    {
        for (int i = 0; ; i ++ )
        {
            int a = primes[i], b = n - a;
            if (!st[b])
            {
                printf("%d = %d + %d\n", n, a, b);
                break;
            }
        }
    }
    return 0;
}

1293. 夏洛克和他的女朋友

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;
const int N = 100010;
int primes[N], cnt;
bool st[N];
int n;

void init(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    cin >> n;
    
    init(n + 1);
    
    if (n <= 2) puts("1");
    else puts("2");
    
    for (int i = 2; i <= n + 1; i ++ )
        if (!st[i]) printf("1 ");
        else printf("2 ");

    return 0;
}

AcWing 196. 质数距离

#include <iostream>
#include <cstring>
#include <algorithm>

typedef long long LL;

using namespace std;
const int N = 1000010;
int primes[N], cnt;
bool st[N];
int l, r;

void init(int n)
{
    memset(st, 0, sizeof st);
    cnt = 0;
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    while (cin >> l >> r)
    {
        init(50010);
        memset(st, 0, sizeof st);

        for (int i = 0; i < cnt; i ++ )
        {
            LL p = primes[i];
            for (LL j = max(p * 2, (l + p - 1) / p * p); j <= r; j += p)
            {
                st[j - l] = true;
            }
        }

        cnt = 0;
        for (int i = 0; i <= r - l; i ++ )
        {
            if (!st[i] && i + l >= 2)
                primes[cnt ++ ] = i + l;
        }

        if (cnt < 2) puts("There are no adjacent primes.");
        else
        {
            int minp = 0, maxp = 0;
            for (int i = 0; i + 1 < cnt; i ++ )
            {
                int d = primes[i + 1] - primes[i];
                if (d < primes[minp + 1] - primes[minp]) minp = i;
                if (d > primes[maxp + 1] - primes[maxp]) maxp = i;
            }

            printf("%d,%d are closest, %d,%d are most distant.\n",
                primes[minp], primes[minp + 1],
                primes[maxp], primes[maxp + 1]);
        }
    }
    return 0;
}

分解质因数

197. 阶乘分解

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int N = 1000010;
int primes[N], cnt;
bool st[N];
int n;

void init(int n)
{
    for (int i = 2; i <= n; i ++ )
    {
        if (!st[i]) primes[cnt ++ ] = i;
        for (int j = 0; primes[j] * i <= n; j ++ )
        {
            st[primes[j] * i] = true;
            if (i % primes[j] == 0) break;
        }
    }
}

int main()
{
    cin >> n;
    
    init(n);
    
    for (int i = 0; i < cnt; i ++ )
    {
        int p = primes[i];
        int s = 0;
        for (int j = n; j; j /= p) s += j / p;
        printf("%d %d\n", p, s);
    }
    
    return 0;    
}

序列的第k个数

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

const int mod = 200907;
typedef long long LL;

LL qmi(int a, int k)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % mod;
        a = (LL)a * a % mod;
        k >>= 1;
    }
    return res;
}

int main()
{
    int T;
    cin >> T;
    while (T -- )
    {
        int a, b, c, k;
        cin >> a >> b >> c >> k;
        if (2 * b == a + c) cout << (a + (b - a) * (LL)(k - 1)) % mod << endl;
        else cout << (LL)a * qmi(b / a, k - 1) % mod << endl;
    }
    return 0;
}

1290. 越狱

#include <iostream>
#include <cstring>
#include <algorithm>

using namespace std;

typedef long long LL;

const int mod = 100003;

int qmi(int a, LL k)
{
    int res = 1;
    while (k)
    {
        if (k & 1) res = (LL)res * a % mod;
        a = (LL)a * a % mod;
        k >>= 1;
    }
    return res;
}

int main()
{
    int m;
    LL n;
    
    cin >> m >> n;
    
    cout << (qmi(m, n) - (LL)m * qmi(m - 1, n - 1) % mod + mod) % mod << endl;
    
    return 0;
}