补题日记CF#749 Div.1+ Div.2
Posted cls1277
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Pro
Codeforces Round #749 (Div. 1 + Div. 2, based on Technocup 2022 Elimination Round 1)
Sol
A. Windblume Ode
分析可得,满足条件的子集大小为n个或n-1个。
所以具体做法为:枚举集合中的每一个数,判断删除后是否可以组成合数,如果可以即为答案。
简单证明:若集合总和为偶数,则必为合数;若总和为奇数,因为奇数=偶数+奇数,则减去一个奇数后必为偶数,即合数。
或者还可以直接判断差值是否为合数。
B. Omkar and Heavenly Tree
思路比较巧妙,思维题。
找b中没有出现过的点做根节点,然后对于每一个查询,都把a和c与根节点相连。
这样就保证了不会有b在a和c的路径上。
C. Omkar and Determination
很容易想到,是否能唯一确定取决于中间是否包含一列不含“关键点”的列,称之为“关键列”。
关键点:上面和左面都是X的点。用vector和bool可以稍微节省一下空间。
然后维护前缀和,对于每个查询只需要看区间内“关键列”的个数是否为0。
注意区间左右端点,因为区间长度至少为2,所以l列不予考虑,直接用sum[r]-sum[l]即可。
Code
A. Windblume Ode
//By cls1277
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<sstream>
#include<set>
#include<cassert>
#include<bitset>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define INF 2147483647
#define eps 1e-7
#define Fo(i,a,b) for(LL i=(a); i<=(b); i++)
#define Ro(i,b,a) for(LL i=(b); i>=(a); i--)
#define Eo(i,x,_) for(LL i=head[x]; i; i=_[i].next)
#define Ms(a,b) memset((a),(b),sizeof(a))
#define lowbit(_) _&(-_)
#define mk(_,__) make_pair(_,__)
#define pii pair<int,int>
#define ls x<<1
#define rs x<<1|1
#define endl '\\n'
inline LL read() {
LL x = 0, f = 1;char c = getchar();
while (!isdigit(c)) { if (c == '-')f = -f;c = getchar(); }
while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48ll), c = getchar();
return x * f;
}
const LL maxn = 105;
int sum,a[maxn];
bool isp(int x) {
Fo(i,2,sqrt(x))
if(x%i==0)
return 0;
return 1;
}
int main() {
ios::sync_with_stdio(false);
#ifdef DEBUG
freopen("data.txt","r",stdin);
#endif
int T; cin>>T;
while(T--) {
int n; cin>>n; sum=0;
Fo(i,1,n) cin>>a[i],sum+=a[i];
if(!isp(sum)) {
cout<<n<<endl;
Fo(i,1,n) cout<<i<<" ";
cout<<endl;
continue;
}
Fo(i,1,n)
if(a[i]%2) {
//if(!isp(sum-a[i])) {
cout<<n-1<<endl;
Fo(j,1,n) {
if(i==j) continue;
cout<<j<<" ";
}
cout<<endl;
break;
}
}
return 0;
}
B. Omkar and Heavenly Tree
//By cls1277
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<sstream>
#include<set>
#include<cassert>
#include<bitset>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define INF 2147483647
#define eps 1e-7
#define Fo(i,a,b) for(LL i=(a); i<=(b); i++)
#define Ro(i,b,a) for(LL i=(b); i>=(a); i--)
#define Eo(i,x,_) for(LL i=head[x]; i; i=_[i].next)
#define Ms(a,b) memset((a),(b),sizeof(a))
#define lowbit(_) _&(-_)
#define mk(_,__) make_pair(_,__)
#define pii pair<int,int>
#define ls x<<1
#define rs x<<1|1
#define endl '\\n'
inline LL read() {
LL x = 0, f = 1;char c = getchar();
while (!isdigit(c)) { if (c == '-')f = -f;c = getchar(); }
while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48ll), c = getchar();
return x * f;
}
const LL maxn = 1e5+5;
int main() {
ios::sync_with_stdio(false);
#ifdef DEBUG
freopen("data.txt","r",stdin);
#endif
int T; cin>>T;
while(T--) {
bitset<maxn>vis;
int n,m,p; cin>>n>>m;
Fo(i,1,m) {
int a,b,c; cin>>a>>b>>c;
vis[b]=1;
}
Fo(i,1,n)
if(!vis[i]) {
p=i;
break;
}
Fo(i,1,n) {
if(i==p) continue;
cout<<p<<" "<<i<<endl;
}
}
return 0;
}
C. Omkar and Determination
//By cls1277
#include<iostream>
#include<cstdio>
#include<queue>
#include<cstring>
#include<string>
#include<algorithm>
#include<cmath>
#include<vector>
#include<map>
#include<stack>
#include<sstream>
#include<set>
#include<cassert>
#include<bitset>
using namespace std;
typedef long long LL;
#define PI acos(-1)
#define INF 2147483647
#define eps 1e-7
#define Fo(i,a,b) for(LL i=(a); i<=(b); i++)
#define Ro(i,b,a) for(LL i=(b); i>=(a); i--)
#define Eo(i,x,_) for(LL i=head[x]; i; i=_[i].next)
#define Ms(a,b) memset((a),(b),sizeof(a))
#define lowbit(_) _&(-_)
#define mk(_,__) make_pair(_,__)
#define pii pair<int,int>
#define ls x<<1
#define rs x<<1|1
#define endl '\\n'
inline LL read() {
LL x = 0, f = 1;char c = getchar();
while (!isdigit(c)) { if (c == '-')f = -f;c = getchar(); }
while (isdigit(c)) x = (x << 1) + (x << 3) + (c ^ 48ll), c = getchar();
return x * f;
}
const LL maxn = 1e6+5;
vector<bool>mp[maxn];
int n,m,sum[maxn];
bitset<maxn>vis;
int main() {
ios::sync_with_stdio(false);
#ifdef DEBUG
freopen("data.txt","r",stdin);
#endif
cin>>n>>m;
Fo(i,1,n)
Fo(j,1,m) {
char x; cin>>x;
if(x=='.') mp[i].push_back(false);
if(x=='X') mp[i].push_back(true);
}
Fo(i,2,n) {
for(int j=1; j<mp[i].size(); j++) {
if(vis[j+1]) continue;
if(mp[i-1][j]==true&&mp[i][j-1]==true) vis[j+1]=1;
}
}
Fo(i,1,m) sum[i]=sum[i-1]+vis[i];
int q; cin>>q;
while(q--) {
int a,b; cin>>a>>b;
if(sum[b]-sum[a]) cout<<"NO"<<endl;
else cout<<"YES"<<endl;
}
return 0;
}
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