So I decided to build an online judge for my graduation project but got rejected immediately by my supervised professor!
Is there any project I can do that is related to CP? :((
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So I decided to build an online judge for my graduation project but got rejected immediately by my supervised professor!
I think the topic is too easy to implement for one year as a final project.
The important thing of such an online code testing system is collecting good problems and test cases for verification.
Implementation of the system itself is quite obvious and easy, and doesn't even take much time. Any senior year students with some programming skills can implement such a system within a month, then will take much more time to fill in problems in the system. So it's more of a labor-intensive topic rather than skill-required topic.
Is there any project I can do that is related to CP? :((
Hi~
To be able to get the hell out of this grey color, i'm learning new data structures. One of those is Fenwick Tree. By copying code of others on the internet, i'm now able to perform some basic operations such as min / max/ sum on a specific interval. Now here come a new problem, how to find the maximum X * f(X) on range [0, r], f(X) here denotes the occurrences of X within that interval. Thank you all, I much appreciate it!
Hi!
Could someone plz explain to me why 1 / 2 modulo 1e9 + 7 = 5e8 + 5? ![ ]()
const int mod = 1e9 + 7;
struct modint {
int n;
modint() :n(0) { ; }
modint(ll m) {
if (m < 0 || mod <= m) {
m %= mod; if (m < 0)m += mod;
}
n = m;
}
operator int() { return n; }
};
bool operator==(modint a, modint b) { return a.n == b.n; }
bool operator<(modint a, modint b) { return a.n < b.n; }
modint operator+=(modint& a, modint b) { a.n += b.n; if (a.n >= mod)a.n -= (int)mod; return a; }
modint operator-=(modint& a, modint b) { a.n -= b.n; if (a.n < 0)a.n += (int)mod; return a; }
modint operator*=(modint& a, modint b) { a.n = ((ll)a.n * b.n) % mod; return a; }
modint operator+(modint a, modint b) { return a += b; }
modint operator-(modint a, modint b) { return a -= b; }
modint operator*(modint a, modint b) { return a *= b; }
modint operator^(modint a, ll n) {
if (n == 0)return modint(1);
modint res = (a * a) ^ (n / 2);
if (n % 2)res = res * a;
return res;
}
ll inv(ll a, ll p) {
return (a == 1 ? 1 : (1 - p * inv(p % a, a)) / a + p);
}
modint operator/(modint a, modint b) { return a * modint(inv(b, mod)); }
modint operator/=(modint& a, modint b) { a = a / b; return a; }
output 5e8 + 4.
Thanks!
The title says it all
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