2018 China Collegiate Programming Contest Final (CCPC-Final 2018)

Problem A. Mischievous Problem Setter

 签到.

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

#define ll long long
#define N 100010
#define pii pair <int, int>
#define x first
#define y second
int t, n, m;
pii a[N];

int main()
{
    scanf("%d", &t);
    for (int kase = 1; kase <= t; ++kase)
    {
        printf("Case %d: ", kase);
        scanf("%d%d", &n, &m);
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i].x);
        for (int i = 1; i <= n; ++i) scanf("%d", &a[i].y);
        sort(a + 1, a + 1 + n);
        int res = 0;
        for (int i = 1; i <= n; ++i) 
        {
            if (a[i].y <= m)
            {
                ++res;
                m -= a[i].y;
            }
            else 
                break;
        }
        printf("%d
", res);
    }
    return 0;
}

Problem K. Mr. Panda and Kakin

题意：

RSA解密

思路：

注意到题目的本意是求$x^(2^{30} + 3) = c pmod (n)$

$从根号处暴力破出p和q，然后求(2^{30} + 3)对 phi(n) = (p - 1) cdot (q - 1) 的逆$

$最后求幂即可，但是注意到大数模乘会爆，所以可以long ;double 或者用中国剩余定理$

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

#define ll long long
int t;
ll p, q;

ll qmod(ll base, ll n, ll MOD)
{
    base %= MOD; 
    ll res = 1; 
    while (n)
    {
        if (n & 1) res = res * base % MOD;
        base = base * base % MOD; 
        n >>= 1;
    }
    return res;
}

void get(ll n)
{
    for (ll i = sqrt(n); i >= 0; --i) if (n % i == 0)  
    {
        p = i;
        q = n / i;   
        return; 
    }
}

ll exgcd(ll a, ll b, ll &x, ll &y)
{
    if (a == 0 && b == 0) return -1;
    if (b == 0) { x = 1, y = 0; return a; } 
    ll d = exgcd(b, a % b, y, x); 
    y -= a / b * x;
    return d; 
}

ll mod_reverse(ll a, ll n)
{
    ll x, y;
    ll d = exgcd(a, n, x, y);
    if (d == 1) return (x % n + n) % n;
    else return -1; 
}

int main()
{
    ll n, c;
    scanf("%d", &t);
    for (int kase = 1; kase <= t; ++kase)
    { 
        printf("Case %d: ", kase);
        scanf("%lld%lld", &n, &c);
        get(n);
        ll r = (p - 1) * (q - 1);
        ll u = mod_reverse(((1ll << 30) + 3), r); 
        ll a = qmod(c, u, p);
        ll b = qmod(c, u, q);     
        b = (b - a + q) % q;
        ll inv = qmod(p, q - 2, q); 
        ll res = b * inv % q;
        res = (res * p % n + a) % n;
        printf("%lld
", res);
    }
    return 0;
}