Codeforces Round #653 (Div. 3) A~E1
减去y，相除， 直接求
--
#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()
int main(int argc, char const *argv[])
{
// #define DEBUG
#ifdef DEBUG
freopen("1.dat","r",stdin);
freopen("ans.dat","w",stdout);
#endif
LL _;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin>>_;
while(_--){
LL a,b,c;
cin>>a>>b>>c;
c-=b;
LL d = c/a;
cout<<(d)*a+b<<endl;
}
return 0;
}
能通过乘以2和除以6达到1的话，那么必然因子只有3和2，而且因子3的个数不小于2的个数
#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()
int main(int argc, char const *argv[])
{
// #define DEBUG
#ifdef DEBUG
freopen("1.dat","r",stdin);
freopen("ans.dat","w",stdout);
#endif
LL _;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin>>_;
while(_--){
LL a;
cin>>a;
// 2和3分别有几个
LL c2 = 0;
LL c3 = 0;
while(a && a%2==0) c2++, a/=2;
while(a && a%3==0) c3++, a/=3;
// 肯定起码有一个三吧
// cout<<c2<<" "<<c3<<" "<<a<<endl;
if(a!=1||c2>c3)
cout<<-1<<endl;
else
cout<<c3*2-c2<<endl;
}
return 0;
}
贪心来做，已经配对的括号肯定不动，然后错位的就只要挪一个就肯定能导致多一个配对的
直接扫一遍就可以
#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()
int main(int argc, char const *argv[])
{
#define DEBUG
#ifdef DEBUG
freopen("1.dat","r",stdin);
freopen("ans.dat","w",stdout);
#endif
LL _;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin>>_;
while(_--){
int n;
cin>>n;
string s;
cin>>s;
//配对的肯定不用管吧
//逆序的就要挪，看有多少是配对的就行
int l = 0;
for(int i=0; i<n; i++){
if(s[i]=='(')
l++;
else
l = max(0, l-1);
}
cout<<l<<endl;
}
return 0;
}
先对所有的数取模，看需要加几，如果一个数需要m次
那么ans起码就要要跳到m*（k-1） + need
#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()
int main(int argc, char const *argv[])
{
// #define DEBUG
#ifdef DEBUG
freopen("1.dat","r",stdin);
freopen("ans.dat","w",stdout);
#endif
LL _;
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
cin>>_;
while(_--){
// 无论做什么，反正都是x要加1的
int n,k;
cin>>n>>k;
int x;
map<int, int> mp;
rep(i,0,n){
cin>>x;
if(x%k==0)
continue;
mp[k-x%k] += 1;
}
LL ans=0;
for(auto e:mp){
ans = max(ans, (LL)(e.se-1)*k+e.fi+1);
}
cout<<ans<<endl;
}
return 0;
}
应该也是贪心吧，思路就是先按时间排序
然后建立两个链表把仅对一个贡献的放到链表里
这里偷懒直接放到优先队列里，遇到一个对两者
都有贡献而且有一个人已经超了的时候看能不能
去掉前面已经选了的来优化结果
#include<bits/stdc++.h>
#include<string.h>
using namespace std;
#define rep(i,j,k) for(LL i=(j); i<(k); ++i)
#define pb push_back
#define PII pair<LL,LL>
#define PLL pair<long long, long long>
#define ini(a,j) memset(a,j,sizeof a)
#define rrep(i,j,k) for(LL i=j; i>=k; --i)
#define fi first
#define se second
#define LL long long
#define beg begin()
#define ed end()
#define all(x) x.begin(),x.end()
struct Book
{
int t;
int al, bo;
Book(){};
Book(int t, int alice, int bob): t(t), al(alice),bo(bob){};
friend bool operator<(const Book& a, const Book& b){
return a.t<b.t;
}
};
int main(int argc, char const *argv[])
{
// #define DEBUG
#ifdef DEBUG
freopen("1.dat","r",stdin);
freopen("ans.dat","w",stdout);
#endif
ios::sync_with_stdio(false);
cin.tie(0);
cout.tie(0);
int a=0, b=0;
int n,k;
cin>>n>>k;
int t, al, bo;
vector<Book> book;
rep(i, 0, n){
cin>>t>>al>>bo;
a += al;
b += bo;
if(!(al||bo))
continue;
book.pb(Book(t, al, bo));
}
if(a<k||b<k){
cout<<-1<<endl;
return 0;
}
priority_queue<Book> quea;
priority_queue<Book> queb;
sort(all(book));
int ans=0;
a=0, b=0;
for(int i=0; i<(int)book.size(); i++){
if(book[i].al&&book[i].bo){
// 不是最多的就直接选吧
if(a>=k&&b>=k){
//尝试能不能换掉？
if(quea.empty()||queb.empty())
break;
int cost=0;
cost += quea.top().t;
quea.pop();
cost += queb.top().t;
queb.pop();
if(book[i].t>=cost)
break;
else{
ans -= cost;
ans += book[i].t;
}
}else if(a>=k||b>=k){
int cost = 0;
if(a>=k&&!quea.empty()){
cost += quea.top().t;
a--;
quea.pop();
}
if(b>=k&&!queb.empty()){
cost += queb.top().t;
b--;
queb.pop();
}
a++, b++;
ans -= cost;
ans += book[i].t;
}else{
a++, b++;
ans += book[i].t;
}
}else if(book[i].al){
if(a>=k)
continue;
a++;
ans += book[i].t;
quea.push(book[i]);
}else{
if(b>=k)
continue;
b++;
ans += book[i].t;
queb.push(book[i]);
}
}
cout<<ans<<endl;
return 0;
}
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一条有梦想的咸鱼
