DMOJ · dmpg18b2
This C++ solution uses mathematical reasoning for DMOJ dmpg18b2 DMPG '18 B2 - Mimi and Modulus. Read the reasoning, inspect the code, or try your own test case below.
Solve first
Read the official problem and plan your algorithm. Reveal the explanation and code when you are ready to compare.
Approach
Mimi and Modulus answer is min(N, M-1), exactly what the code prints.
Problem and code
Written by benbenyaojifen. Try the problem first, then compare your approach with the code.
#include <bits/stdc++.h>
using namespace std;
using ll = long long;
using i128 = __int128;
const int inf = 1e9;
const ll INF = 1e18; //❄️
int main() {
ios::sync_with_stdio(0); cin.tie(0);
int n, m; cin >> n >> m;
cout << min(n, m - 1) << '\n';
return 0;
}
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