标签:std 插入 detail abc name namespace tco min 解法
先找到循环节的长度乘以每个循环节中\(a\) 的数量,然后在加上
注意 long long
#include <bits/stdc++.h>
using namespace std;
int main() {
long long n,a,b;
cin >> n >> a >> b;
cout <<( n / (a + b) * a + min(a,n % (a + b)) );
return 0;
}
枚举
\(O(1)\) 解法 先算出max(ceil(a/0.08),ceil(b/0.1)),再看这个数能否回去
#include <bits/stdc++.h>
using namespace std;
int main() {
int a,b;
cin >> a >> b;
for(int i = 1;i <= 1250; ++i) {
if(floor(i * 0.08) == a && floor(i * 0.1) == b) {
cout << i ;
return 0;
}
}
cout << -1;
return 0;
}
加个反转标记,deque的插入删除都是\(O(1)\)
https://blog.csdn.net/like_that/article/details/98446479
#include <bits/stdc++.h>
using namespace std;
int main() {
ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
deque<char> q;
string s;
int Q,t,f;
bool k = 1;
char c;
cin >> s >> Q;
for(int i = 0;i < s.size(); ++i) {
q.push_back(s[i]);
}
while(Q --) {
cin >> t;
if(t == 1) k = !k;
else {
cin >> f >> c;
if(f == 1) {
if(k == 1) q.push_front(c);
else q.push_back(c);
}
else {
if(k == 1) q.push_back(c);
else q.push_front(c);
}
}
}
if(k) for(int i = 0;i < q.size(); ++i) cout << q[i] ;
else for(int i = q.size() - 1;i >= 0; --i) cout << q[i];
return 0;
}
AtCoder Beginner Contest 158 BCD
标签:std 插入 detail abc name namespace tco min 解法
原文地址:https://www.cnblogs.com/lukelmouse/p/12489964.html