智算之道2020复赛参考代码

T1

#include <cstdio>
#define

LL pow10(LL m)
{
    return m == 0 ? 1 : 10 * pow10(m - 1);
}

int main()
{
    LL aa, b;
    scanf("%lld%lld", &aa, &b);
    LL len = 0;
    LL tmp = aa;
    while(tmp)
    {
        len++;
        tmp /= 10;
    }
    LL result = 0;
    for(LL i = 100; i < 1000; i++)
    {
        LL a = i * pow10(len) + aa;
        if(a % b == 0)
        {
            result++;
        }
    }
    printf("%lld",result);
    return 0;
}

T2

#include<cstdio>
#include <algorithm>
using namespace std;

#define
#define

typedef long long LL;
const int N = 2020;

int n, k, w1, w2;
pair<int, int> p[N];
int f[N];

int main()
{
    //freopen("T2.in", "r", stdin);

    scanf("%d%d%d%d", &n, &k, &w1, &w2);
    int m = n << 1;
    if(w1 * 2 <= w2)
    {
        return 0 * printf("%lld\n", (LL)m * w1);
    }

    for(int i = 0; i < k; i ++)
    {
        scanf("%d%d", &p[i].x, &p[i].y);
    }

    sort(p, p + k);

    int cnt = 0;
    for(int i = 0; i < k; i ++)
    {
        f[i] = 1;
        for(int j = 0; j < i; j ++)
        {
            if(p[j].x < p[i].x and p[j].y < p[i].y and f[j] + 1 > f[i])  f[i] = f[j] + 1;
        }

        if(f[i] > cnt)
        {
            cnt = f[i];
        }
    }

    m -= cnt * 2;

    printf("%lld\n", (LL)m * w1 + (LL)cnt * w2);

    return 0;
}

T3

#include<iostream>
#include<vector>
using namespace std;

typedef unsigned long long LL;

vector<pair<int,int> >vec;

int main()
{
    //freopen("T3.in", "r", stdin);

    LL n,k;
    scanf("%llu%llu",&k,&n);

    if(n == 2 && k == 1)
    {
        printf("2 1\n");
        printf("1 2\n");
        return 0;
    }
    else if(n == 4 && k == 3)
    {
        printf("4 5\n");
        printf("1 2\n2 3\n3 4\n1 3\n2 4\n");
        return 0;
    }

    LL tmp = k;
    int cnt = 0;
    while(tmp) //k有几位二进制位
    {
        cnt++;
        tmp /= 2;
    }

    int N = cnt + 2;
    for(int i = 2; i <= cnt + 1; i++)
    {
        for(int j = i + 1; j <= cnt + 2; j++)
        {
            vec.push_back({i,j});
        }
    }

    for(int i = 0; i < cnt; i++)
    {
        if((k >> i) & 1)
        {
            int nex = cnt - i + 1;
            vec.push_back({1,nex});
        }
    }

    printf("%d %d\n", N, (int)vec.size());
    for(int i = 0; i < vec.size(); i++)
    {
        printf("%d %d\n",vec[i].first,vec[i].second);
    }

    return 0;
}

T4

#include<iostream>
using namespace std;

typedef long long LL;

const int maxn = 8e7 + 7;
const LL mod = 1ll << 32;

int prime[5000000], cnt;
int primePos[maxn];

//线性筛选法求质数
void getprime()
{
    for(int num = 2; num < maxn; num++)
    {
        if(primePos[num] == 0)//为0时才是质数
        {
            primePos[num] = num;
            prime[++cnt] = num;
        }

        for(int j = 1; j <= cnt && num * prime[j] < maxn; j++)
        {
            primePos[num * prime[j]] = 1;
            if(num % prime[j] == 0)
            {
                break;
            }
        }
    }
}

int main()
{
    //freopen("T4.in", "r", stdin);

    getprime();

    LL n, A, B;
    scanf("%lld%lld%lld", &n, &A, &B);

    for(int num = 2; num <= n; num++)
    {
        primePos[num] = 0;
    }

    for(int i = 1; i <= cnt; i++)
    {
        LL now = prime[i];
        while(now <= n)
        {
            primePos[now] = i;
            now = now * prime[i];
        }
    }

    for(int num = 2; num <= n; num++)
    {
        if(primePos[num])
        {
            A = (A * prime[primePos[num]] % mod + B) % mod;
        }
    }

    printf("%lld\n", A);

    return 0;
}