Codeforces Round #662 (Div. 2) 题解
Codeforces Round #662 (Div. 2) 题解
-
阅读题,题意比较难理解,但是读懂题多画几个例子之后就能发现答案其实就是 ans=n/2+1;
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
int t,n;
int main()
{
ios::sync_with_stdio(false);
cin.tie(0);cout.tie(0);
cin>>t;
while(t--){
cin>>n;
int ans;
if(n&1){
ans=n/2+1;
} else {
ans=n/2+1;
}
cout<<ans<<"\n";
}
return 0;
}
- B. Applejack and Storages
- 统计一下个数,顺带记录一下2-4个的,4-6个的,6-8个的和8个以上的就行了
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
int t, n;
const int N = 1e5 + 10;
int a[N];
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
//cin >> t;
//while (t--) {
cin >> n;
int n2 = 0, n4 = 0, n6=0,n8=0;
memset(a, 0, sizeof(a));
for (int i = 0; i < n; ++i) {
int tmp;
cin >> tmp;
a[tmp]++;
if (a[tmp] == 2) n2++;
if (a[tmp] == 4) n2--, n4++;
if(a[tmp]==6) n4--,n6++;
if(a[tmp]==8) n6--,n8++;
}
int q;
cin >> q;
//cout<<q<<"\n";
for (int i = 0; i < q; ++i) {
char op; int num;
cin >> op >> num;
if (op == '+') {
a[num]++;
if (a[num] == 2) n2++;
if (a[num] == 4) n4++, n2--;
if(a[num]==6) n4--,n6++;
if(a[num]==8) n6--,n8++;
}
else {
a[num]--;
if(a[num]==7) n8--,n6++;
if(a[num]==5) n6--,n4++;
if (a[num] == 3) n4--, n2++;
if (a[num] == 1) n2--;
}
if(n8>=1){
cout<<"YES\n";
continue;
}
if(n6>=1){
if(n6==1){
if(n4>=1 ||n2>=1){
cout<<"YES\n";
continue;
}else{
cout<<"NO\n";
continue;
}
}else{
cout<<"YES\n";
continue;
}
}else{
if(n4>=1){
if(n4==1){
if(n2>=2){
cout<<"YES\n";
continue;
}else {
cout<<"NO\n";
continue;
}
}else{
cout<<"YES\n";
continue;
}
}else{
cout<<"NO\n";
continue;
}
}
}
//}
return 0;
}
-
记录下个数,找到个数最多的几个,然后先排个数最多的那几个,计算下最大的最小距离就行了(比赛的时候傻逼了用二分去搜索答案,又是二分出锅qwq
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
typedef long double ld;
const int N=1e5+10;
int t,n;
int a[N];
struct node{
int id,num;
}p[N];
bool cmp(node a,node b)
{
return a.num>b.num;
}
bool check(int x,int mn,int n1) //n1 mam
{
if(x*(n1-1)+mn*(n1)<=n) return 1;
else return 0;
}
int main()
{
ios::sync_with_stdio(false);
cin.tie(0); cout.tie(0);
cin>>t;
while(t--){
cin>>n;
memset(a,0,sizeof(int)*(n+1));
int mam=0,mnum=0;
for(int i=1;i<=n;++i){
int tmp;
cin>>tmp;
a[tmp]++;
mam=max(mam,a[tmp]);
}
for(int i=0;i<=n;++i){
if(a[i]==mam) mnum++;
}
int ans=(n-mnum*mam)/(mam-1)+mnum-1;
cout<<ans<<"\n";
}
return 0;
}
DEF明天起来补叭