目录
[A. Rook](#A. Rook)
[B. YetnotherrokenKeoard](#B. YetnotherrokenKeoard)
[C. Removal of Unattractive Pairs](#C. Removal of Unattractive Pairs)
[D. Jumping Through Segments](#D. Jumping Through Segments)
[E. Good Triples](#E. Good Triples)
A. Rook
问题分析:
给一个棋子将其同行同列的位置输出
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
const int N =1e5+7;
signed main()
{IOS
use{
string a;cin>>a;
char x=a[0];
int num=a[1]-'0';
for(int i=1;i<=8;i++){
if(i!=num)
cout<<x<<i<<endl;
}
for(int i=0;i<=7;i++){
if('a'+i!=x){
cout<<(char)('a'+i)<<num<<endl;
}
}
}
return 0;
}
B. YetnotherrokenKeoard
问题分析:
一个字符串,当当前字符为'b'时,删除新字符串最右边的 小写字符,当当前字符为'B'时删除新字符串最右边的大写字符,没有则不删除,除去'b'和'B'的其他字符添加到新字符串当中,然后输出。
可以通过栈来存储新字符串当中的大小写字符的位置,然后遇到相应的'b' or 'B',将栈顶的位置标记,最后按照标记输出原字符串的字符即可
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
const int N =1e5+7;
signed main()
{IOS
use{
string a;cin>>a;
stack<int>lowi,upi;
vct<bool>st(a.LEN);
for(int i=0;i<a.LEN;i++){
if(a[i]=='b'){
st[i]=1;
if(lowi.empty())continue;
else st[lowi.top()]=1,lowi.pop();
}
else if(a[i]=='B'){
st[i]=1;
if(upi.empty())continue;
else st[upi.top()]=1,upi.pop();
}
else {
if(islower(a[i])){
lowi.push(i);
}else {
upi.push(i);
}
}
}
for(int i=0;i<a.LEN;i++){
if(!st[i])cout<<a[i];
}cout<<"\n";
}
return 0;
}
C. Removal of Unattractive Pairs
问题分析:
给定字符串,如果相邻两个字符不同,可以删除这两个字符,求出最终可以剩下几个字符
我们统计每个字符的数量,为了保证最后剩下的字符个数最少,我们贪心地用个数次大的字符与个数最大的字符相抵消,因为需要维护数量的降序排序,我们选择使用大根堆来维护,我们每次弹出堆顶的两个元素,将他们互相抵消,也就是都减一,如果大于零就再放回堆当中,最终堆内会只剩下一个数,也就是最短的字符串长度
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
const int N =1e5+7;
signed main()
{IOS
use{
int n;cin>>n;
string a;cin>>a;
vct<int>cnt(26);
for(int i=0;i<a.LEN;i++){
cnt[a[i]-'a']++;
}
priority_queue<int>q;
for(int i=0;i<26;i++){
if(cnt[i]!=0)q.push(cnt[i]);
}
while(q.size()>1){
int x=q.top();
q.pop();
int y=q.top();
q.pop();
x--;
y--;
if(x>0)q.push(x);
if(y>0)q.push(y);
}
if(q.empty())cout<<"0"<<endl;
else cout<<q.top()<<endl;
}
return 0;
}
D. Jumping Through Segments
问题分析:
n个线段,每次可以选择走0~k格,第次一定要落在第个线段内部,要求在满足条件的情况下使得k最小.
对于k来说,k的大小对于结果的影响是单调的,故可以用二分答案,我们只需要确定第步走出的范围能够与第个线段有交集即可
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
const int N =1e5+7;
signed main()
{
IOS
use{
int n;
cin >> n;
int ans;
vct<int> l(n + 1);
vct<int> r(n + 1);
for (int i = 1; i <= n; i++) cin >> l[i] >> r[i];
int L = 0, R = 1e9;
int mid;
auto check = [&](int mid) -> bool {
int dis1 = 0, dis2 = 0;
for (int i = 1; i <= n; i++) {
dis1 -= mid, dis2 += mid;
if (dis1 > r[i] || dis2 < l[i]) return false;
mmax(dis1, l[i]), mmin(dis2, r[i]);
}
return true;
};
while (L <= R) {
mid = (L + R) >> 1;
if (check(mid)) {
R = mid - 1;
ans = mid;
} else L = mid + 1;
}
cout << ans << endl;
}
return 0;
}
E. Good Triples
问题分析:
给定整数n,求出有多少个三元组满足题目条件,且若三元组的元素相同但排序不同,则 算是不同的三元组。
我们通过样例以及向下推导可知:
故满足的公式,且为n各位数字得到结果的乘积
cpp
#include<bits/stdc++.h>
#pragma GCC optimize("Ofast")
#define INF 0x3f3f3f3f
#define IOS ios::sync_with_stdio(false);cin.tie(0);
#define int long long
#define pb push_back
#define vct vector
#define checkbit __builtin_popcount
#define gcd __gcd
#define use int T;cin>>T;while(T--)
#define LEN length()
#define all(a) a.begin(),a.end()
template<class T> bool mmax(T &u, T v) { return u < v ? (u = v, 1) : 0; }
template<class T> bool mmin(T &u, T v) { return u > v ? (u = v, 1) : 0; }
#define lowbit(x) (x&(-x))
#define yes cout<<"YES"<<endl
#define no cout<<"NO"<<endl
using namespace std;
typedef pair<int,int>pii;
const int N =1e5+7;
signed main()
{
IOS
use{
string a;cin>>a;
int ans=1;
for(int i=0;i<a.LEN;i++){
ans*=(a[i]-'0'+2)*(a[i]-'0'+1)/2;
}cout<<ans<<endl;
}
return 0;
}