Roman and Numbers time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard output Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change his…
题目链接: http://codeforces.com/problemset/problem/401/D D. Roman and Numbers time limit per test4 secondsmemory limit per test512 megabytes 问题描述 Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sere…
D. Roman and Numbers time limit per test 4 seconds memory limit per test 512 megabytes input standard input output standard output Roman is a young mathematician, very famous in Uzhland. Unfortunately, Sereja doesn't think so. To make Sereja change h…
D - Jzzhu and Numbers 这个容斥没想出来... 我好菜啊.. f[ S ] 表示若干个数 & 的值 & S == S得 方案数, 然后用这个去容斥. 求f[ S ] 需要用SOSdp #include<bits/stdc++.h> #define LL long long #define fi first #define se second #define mk make_pair #define PLL pair<LL, LL> #define…
A. Vanya and Cards time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Vanya loves playing. He even has a special set of cards to play with. Each card has a single integer. The number on the ca…
A. Mahmoud and Longest Uncommon Subsequence time limit per test 2 seconds memory limit per test 256 megabytes input standard input output standard output While Mahmoud and Ehab were practicing for IOI, they found a problem which name was Longest comm…
D. Roman Digits time limit per test 1 second memory limit per test 256 megabytes input standard input output standard output Let's introduce a number system which is based on a roman digits. There are digits I, V, X, L which correspond to the numbers…
链接:https://codeforces.com/contest/1100/problem/A 题意: 给定n,k. 给定一串由正负1组成的数. 任选b,c = b + i*k(i为任意整数).将c所有c位置的数删除,求-1和1个数差值绝对值的最大值. 思路: 暴力遍历 代码: #include <bits/stdc++.h> using namespace std; int a[110]; int main() { int n,k; scanf("%d%d",&…
#include <iostream> #include <algorithm> using namespace std; int main(){ int n,m; cin >> n >> m; ) || m > (n + )* ) cout<< -; else if( n>= m) { ; i < m; ++ i) cout<<"; "; }>=m){ ; i< m - n; ++…
