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

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

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

 ll exgcd(ll a, ll b, ll &x, ll &y)
 {
     if (a ==  && b == ) return -;
     if (b == ) { x = , y = ; 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 == ) return (x % n + n) % n;
     else return -;
 }

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