HDU Math Problems

1576

const int mod = 9973;

n = a - a / mod * mod;

a / b = ans;

ans * b = a = a / mod * mod + n;

n = b * ans - a / mod * mod;

n = b * ans + mod * y;

extended_gcd(b, mod, ans, y);

#define PRON "hdu1576"
#define LL "%lld"
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int MOD = 9973;

int Tcase;

ll extended_gcd(ll a, ll b, ll & x, ll & y){
    if (b == 0){
        x = 1, y = 0;
        return a;
    }

    ll d = extended_gcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - a / b * y;
    
    return d;
}

int main(){
#ifndef ONLINE_JUDGE
    freopen(PRON ".in", "r", stdin);
#endif

    ll a, b, x, y;

    scanf("%d", &Tcase);
    while (Tcase --){
        scanf(LL LL, &a, &b);
        extended_gcd(b, MOD, x, y);
        x = ((x * a % MOD) + MOD) % MOD;
        printf(LL "
", x);
    }
}

  

2824

Σphi[i]

#define PRON "hdu2824"
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef long long ll;

const int MAXN = 3000000;

int n, m;
ll phi[MAXN + 10];

void get_phi(){
    memset(phi, 0, sizeof phi);
    phi[1] = 1;
    for (int i = 2; i <= MAXN; i ++)
        if (!phi[i]){
            for (int j = i; j <= MAXN; j += i){
                if (!phi[j])
                    phi[j] = j;
                phi[j] = phi[j] / i * (i - 1);
            }
        }
}

int main(){
#ifndef ONLINE_JUDGE
    freopen(PRON ".in", "r", stdin);
#endif

    get_phi();
    for (int i = 1; i <= MAXN; i ++)
        phi[i] += phi[i - 1];

    while (scanf("%d %d", &n, &m) == 2)
        cout << phi[m] - phi[n - 1] << endl;
}

1573

中国剩余定理的一般形式

#define PRON "hdu1573"
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef int ll;

const int MAXN = 10 + 5;

int Tcase, _max, n, a[MAXN], b[MAXN];

ll extended_gcd(ll a, ll b, ll & x, ll & y){
    if (b == 0){
        x = 1, y = 0;
        return a;
    }

    ll d = extended_gcd(b, a % b, x, y);
    ll temp = x;
    x = y; 
    y = temp - a / b * y;

    return d;
}

ll normal_crt(){
    ll m1, m2, r1, r2, x, y;

    //solve N = r1 (mod m1)
    //      N = r2 (mod m2)
    m1 = a[0], r1 = b[0];
    for (int i = 1; i < n; i ++){
        m2 = a[i], r2 = b[i];
    
        //solve d = x * m1 + y * m2
        //(x, y) is the solution to the equation above
        //solve c = r2 - r1 = y * m2 - x * m1
        //(x0, y0) is the solution to the equation above
        //x0 = x * c / d, y0 = x * c / d
        ll d = extended_gcd(m1, m2, x, y);
        ll c = r2 - r1;
        if (c % d)
            return 0;

        ll t = m2 / d;
        x = (x * c / d % t + t) % t;
        
        //r1 is the solution to the equaions from 1st to ith
        r1 += m1 * x;
        //m1 is the lcm of m1 to mi
        m1 *= t;
    }

    if (_max < r1)
        return 0;

    //if (x0, y0) is one of the solution
    //(x0 + k * m2 / d, y0 - k * m1 / d) (k -> Z) also apply
    return (_max - r1) / m1 + 1 - (bool)(r1 == 0);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen(PRON ".in", "r", stdin);
#endif

    scanf("%d", &Tcase);
    while (Tcase --){
        scanf("%d %d", &_max, &n);
        for (int i = 0; i < n; i ++)
            scanf("%d", a + i);
        for (int i = 0; i < n; i ++)
            scanf("%d", b + i);

        printf("%d
", normal_crt());
    }
}

1370

中国剩余定理 

#define PRON "hdu1370"
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;
typedef int ll;

const int MAXN = 10;
const int MOD = 21252;

int Tcase, cnt, st, a[MAXN], b[MAXN];

ll extended_gcd(ll a, ll b, ll & x, ll & y){
    if (b == 0){
        x = 1, y = 0;
        return a;
    }

    ll d = extended_gcd(b, a % b, x, y);
    ll temp = x;
    x = y;
    y = temp - a / b * y;

    return d;
}

ll inv(ll a, ll n){
    ll x, y;
    ll d = extended_gcd(a, n, x, y);
    return d == 1 ? (x + n) % n : -1;
}

ll crt(int n){
    ll ret = 0, m = 1;
    
    for (int i = 0; i < n; i ++)
        a[i] %= b[i], m *= b[i]; 

    for (int i = 0; i < n; i ++)
        ret = (ret + a[i] * (m / b[i]) * inv(m / b[i], b[i])) % m;

    ret -= st;
    return ret + MOD * (bool)(ret <= 0);
}

int main(){
#ifndef ONLINE_JUDGE
    freopen(PRON ".in", "r", stdin);
#endif

    cnt = 0;
    b[0] = 23, b[1] = 28, b[2] = 33;

    scanf("%d", &Tcase);
    while (scanf("%d %d %d %d", &a[0], &a[1], &a[2], &st) == 4 && !(a[0] == -1 && a[1] == -1 && a[2] == -1))
        printf("Case %d: the next triple peak occurs in %d days.
", ++ cnt, crt(3));
}