题目链接
好不太容易AK，不能不写题解。
A-第k小数
做法：nth_element的应用，去年寒假牛客训练营出过一次。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=5e6+10;
int a[N],n,k;
int main()
{
int _;cin>>_;while(_--)
{
n=read(),k=read();
rep(i,0,n-1) a[i]=read();
nth_element(a,a+k-1,a+n);
printf("%d\n",a[k-1]);
}
}
B-不平行的直线
做法:将所有直线斜率求出，然后去重即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
int gcd(int a,int b) { return b?gcd(b,a%b):a;}
vector<pair<int,int> >G;
int n;
const int N=2e2+10;
int x[N],y[N];
int main()
{
n=read();
rep(i,1,n) x[i]=read(),y[i]=read();
rep(i,1,n)
{
rep(j,i+1,n){
int fz=y[i]-y[j],fm=x[i]-x[j];
int gc=gcd(fz,fm);
fz/=gc,fm/=gc;
if(fm<0) fz=-fz,fm=-fm;
G.push_back(make_pair(fz,fm));
}
}
sort(G.begin(),G.end());
G.erase(unique(G.begin(),G.end()),G.end());
printf("%d\n",G.size());
}
C-丢手绢
做法:将原数组扩大两倍，然后同时枚举起点、终点，二分找尽量对两端最远的点即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=2e5+10;
int n;
ll a[N],sum[N],x[N];
int main()
{
n=read();
//++n;
rep(i,1,n) a[i]=read();
for(int i=1,j=n+1;i<=n;++i,++j) a[j]=a[i];
//rep(i,1,2*n) printf("%d ",a[i]);
//puts("");
//
//n=2*n;
rep(i,1,2*n) sum[i]=sum[i-1]+a[i];
int id1=1,id2=n+1;
ll ans=0;
while(id1<=n)
{
int l=id1,r=id2-1;
while(l<=r)
{
int mid=l+r>>1;
ll t1=sum[id2-1]-sum[mid];
ll t2=sum[mid]-sum[id1-1];
//printf("id1:%d id2:%d mid:%d t1:%lld t2:%lld\n",id1,id2,mid,t1,t2);
if(t1>t2){
ans=max(ans,t2);
l=mid+1;
}
else{
r=mid-1;
ans=max(ans,t1);
}
}
++id1,++id2;
}
printf("%lld\n",ans);
}D-二分
做法：枚举当前这个人说的是真话。如果是+ 那就定这个答案数 是 v-1.
如果当前这个人是- 那就定这个答案数 是 v+1
已经知道了最初的数了，就对+号和-号分开判断 简单的二分判断有多少个是正确的即可。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e5+10;
int X[N],Y[N],lx,ly,n;
map<int,int>mp;
struct node
{
int v;
char s[2];
}a[N];
int cal(int v)
{
int res=mp[v],id;
id=upper_bound(X+1,X+1+lx,v)-X;
//printf("v:%d id:%d %d\n",v,id,lx-id+1);
res+=lx-id+1;
id=lower_bound(Y+1,Y+1+ly,v)-Y;
//printf("v:%d id:%d %d\n",v,id,id-1);
res+=id-1;
//puts("");
return res;
}
int main()
{
n=read();
rep(i,1,n)
{
a[i].v=read();cin>>a[i].s;
if(a[i].s[0]=='.')mp[a[i].v]++;
else if(a[i].s[0]=='+') X[++lx]=a[i].v;
else if(a[i].s[0]=='-') Y[++ly]=a[i].v;
}
sort(X+1,X+1+lx);
sort(Y+1,Y+1+ly);
int ans=max(ans,max(lx,ly));
rep(i,1,n)
{
int res;
if(a[i].s[0]=='.') res=cal(a[i].v);
else if(a[i].s[0]=='+') res=cal(a[i].v-1);
else res=cal(a[i].v+1);
ans=max(ans,res);
//printf("v:%d res:%d\n",a[i].v,res);
}
printf("%d\n",ans);
}
/*
5
0 +
1 -
2 -
3 -
4 +
*/
E-交换
做法：置换群的水题了。
#pragma GCC optimize(2)
#include<bits/stdc++.h>
#define ll long long
#define maxn 1005
#define inf 1e9
#define pb push_back
#define rep(i,a,b) for(int i=a;i<=b;i++)
#define per(i,a,b) for(int i=a;i>=b;i--)
using namespace std;
inline ll read()
{
ll x=0,w=1; char c=getchar();
while(c<'0'||c>'9') {if(c=='-') w=-1; c=getchar();}
while(c<='9'&&c>='0') {x=(x<<1)+(x<<3)+c-'0'; c=getchar();}
return w==1?x:-x;
}
const int N=1e5+10;
int n,fa[N],vis[N];
struct node
{
int v,id;
}a[N];
bool cmp(node a,node b)
{
return a.v<b.v;
}
int bfs()
{
int ans=0;
rep(i,1,n)
{
if(vis[i]) continue;
int id=i,len=0;
vis[id]=1;
while(vis[fa[id]]==0) {
id=fa[id];
vis[id]=1;
len++;
}
ans+=len;
}
return ans;
}
int main()
{
n=read();
rep(i,1,n){
a[i].v=read();
a[i].id=i;
}
sort(a+1,a+1+n,cmp);
rep(i,1,n) fa[a[i].id]=i;
printf("%d\n",bfs());
}