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

Problem A. Mischievous Problem Setter

 签到.

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 #define N 100010
 6 #define pii pair <int, int>
 7 #define x first
 8 #define y second
 9 int t, n, m;
10 pii a[N];
11 
12 int main()
13 {
14     scanf("%d", &t);
15     for (int kase = 1; kase <= t; ++kase)
16     {
17         printf("Case %d: ", kase);
18         scanf("%d%d", &n, &m);
19         for (int i = 1; i <= n; ++i) scanf("%d", &a[i].x);
20         for (int i = 1; i <= n; ++i) scanf("%d", &a[i].y);
21         sort(a + 1, a + 1 + n);
22         int res = 0;
23         for (int i = 1; i <= n; ++i) 
24         {
25             if (a[i].y <= m)
26             {
27                 ++res;
28                 m -= a[i].y;
29             }
30             else 
31                 break;
32         }
33         printf("%d\n", res);
34     }
35     return 0;
36 }

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Problem K. Mr. Panda and Kakin

题意：

RSA解密

思路：

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

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

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

 1 #include <bits/stdc++.h>
 2 using namespace std;
 3 
 4 #define ll long long
 5 int t;
 6 ll p, q;
 7 
 8 ll qmod(ll base, ll n, ll MOD)
 9 {
10     base %= MOD; 
11     ll res = 1; 
12     while (n)
13     {
14         if (n & 1) res = res * base % MOD;
15         base = base * base % MOD; 
16         n >>= 1;
17     }
18     return res;
19 }
20 
21 void get(ll n)
22 {
23     for (ll i = sqrt(n); i >= 0; --i) if (n % i == 0)  
24     {
25         p = i;
26         q = n / i;   
27         return; 
28     }
29 }
30 
31 ll exgcd(ll a, ll b, ll &x, ll &y)
32 {
33     if (a == 0 && b == 0) return -1;
34     if (b == 0) { x = 1, y = 0; return a; } 
35     ll d = exgcd(b, a % b, y, x); 
36     y -= a / b * x;
37     return d; 
38 }
39 
40 ll mod_reverse(ll a, ll n)
41 {
42     ll x, y;
43     ll d = exgcd(a, n, x, y);
44     if (d == 1) return (x % n + n) % n;
45     else return -1; 
46 }
47 
48 int main()
49 {
50     ll n, c;
51     scanf("%d", &t);
52     for (int kase = 1; kase <= t; ++kase)
53     { 
54         printf("Case %d: ", kase);
55         scanf("%lld%lld", &n, &c);
56         get(n);
57         ll r = (p - 1) * (q - 1);
58         ll u = mod_reverse(((1ll << 30) + 3), r); 
59         ll a = qmod(c, u, p);
60         ll b = qmod(c, u, q);     
61         b = (b - a + q) % q;
62         ll inv = qmod(p, q - 2, q); 
63         ll res = b * inv % q;
64         res = (res * p % n + a) % n;
65         printf("%lld\n", res);
66     }
67     return 0;
68 }

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