A. Trip For Meal
给定一个三角形，你要在三角形上不断地走n步，可以走回头路，问你最短路是多少
贪心地走当前能走的最短边即可，模拟一百能过
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
char ch=getchar();int s=0,w=1;
while(ch<48||ch>57){if(ch==‘-‘)w=-1;ch=getchar();}
while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
return s*w;
}
inline void write(int x){
if(x<0)putchar(‘-‘),x=-x;
if(x>9)write(x/10);
putchar(x%10+48);
}
int gcd(int a, int b){
if(a==0) return b;
if(b==0) return a;
if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
else if(!(b&1)) return gcd(a,b>>1);
else if(!(a&1)) return gcd(a>>1,b);
else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}
int n,a,b,c;
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin>>n>>a>>b>>c;
n--;
int ans=0,now=1;
while(n--){
if(now==1){
if(a<b) ans+=a,now=2;
else ans+=b,now=3;
}
else if(now==2){
if(a<c) ans+=a,now=1;
else ans+=c,now=3;
}
else if(now==3){
if(b<c) ans+=b,now=1;
else ans+=c,now=2;
}
}
cout<<ans;
return 0;
}
B. Yet Another Array Partitioning Task
给你一个序列，你要把它分成k段，每段的前m大数字会被记入答案的贡献当中，问你答案最大是多少
首先考虑前m*k大个数字应该都是要被选中的，答案应该就是他们的和
比较狗屎的是要输出分段的位置，我是用一个map存哪些数字要被输出，每输出了m个数字就分一次段
注意判断一下只输出k段
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
char ch=getchar();int s=0,w=1;
while(ch<48||ch>57){if(ch==‘-‘)w=-1;ch=getchar();}
while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
return s*w;
}
inline void write(int x){
if(x<0)putchar(‘-‘),x=-x;
if(x>9)write(x/10);
putchar(x%10+48);
}
int gcd(int a, int b){
if(a==0) return b;
if(b==0) return a;
if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
else if(!(b&1)) return gcd(a,b>>1);
else if(!(a&1)) return gcd(a>>1,b);
else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}
int n,m,k,a[maxn],b[maxn];
mpii mp;
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin>>n>>m>>k;re(i,1,n) cin>>a[i],b[i]=a[i];
sort(b+1,b+1+n,greater<int>());
int sum=0;
re(i,1,m*k) sum+=b[i],mp[b[i]]++;
cout<<sum<<endl;
int cnt=0,can=0;
re(i,1,n){
if(mp[a[i]]!=0){
cnt++;
mp[a[i]]--;
if(cnt==m){
cnt=0;
cout<<i<<‘ ‘;
can++;
if(can==k-1) break;
}
}
}
return 0;
}
C. Do you want a date?
对一个序列，他的每个子串的最大值减去最小值都要被计入答案，求答案
考虑把序列由小到大排序，然后看每个数字被加了几次或者减了几次
显然，某个数字作为最大值的情况是在他左边的数字任选，右边的数字一个都不选，作为最小值时相反
那么答案就是2的左边数字或者右边数字次幂，注意取模
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=1e9+7;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now){}
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
char ch=getchar();int s=0,w=1;
while(ch<48||ch>57){if(ch==‘-‘)w=-1;ch=getchar();}
while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
return s*w;
}
inline void write(int x){
if(x<0)putchar(‘-‘),x=-x;
if(x>9)write(x/10);
putchar(x%10+48);
}
int gcd(int a, int b){
if(a==0) return b;
if(b==0) return a;
if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
else if(!(b&1)) return gcd(a,b>>1);
else if(!(a&1)) return gcd(a>>1,b);
else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}
int n,x[maxn];
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin>>n;re(i,1,n) cin>>x[i];
sort(x+1,x+1+n);
int ans=0;
re(i,1,n){
ans=(ans+qpow(2,i-1)*(x[i]%MOD))%MOD;
ans=(ans+MOD-qpow(2,n-i)*(x[i]%MOD))%MOD;
}
cout<<ans;
return 0;
}
D. Coffee and Coursework (Hard Version)
你有n杯咖啡，每天可以喝任意多杯咖啡，每杯咖啡有一个贡献，但是喝的咖啡过多会有损益
喝第一杯咖啡损益为0，第二杯为1，第三杯以此类推
进入下一天时损益清零
你需要找到最少多少天可以使贡献大于等于一个给定的值
首先把贡献从大到小排序
考虑二分天数，在check答案时应该这样考虑，比如你有三天时间去完成任务，那么就让最大的三杯咖啡损益都为0
这样一定能找到当前天数下最多能达到多少贡献
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=100000007;
const db eps=1e-10;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now);
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
char ch=getchar();int s=0,w=1;
while(ch<48||ch>57){if(ch==‘-‘)w=-1;ch=getchar();}
while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
return s*w;
}
inline void write(int x){
if(x<0)putchar(‘-‘),x=-x;
if(x>9)write(x/10);
putchar(x%10+48);
}
int gcd(int a, int b){
if(a==0) return b;
if(b==0) return a;
if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
else if(!(b&1)) return gcd(a,b>>1);
else if(!(a&1)) return gcd(a>>1,b);
else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}
int n,m,a[maxn];
int sum[maxn],del[maxn];
bool chk(int now){
re(i,1,now) del[i]=0;
int cnt=0;
re(i,1,n){
cnt+=a[i]-(i-1)/now;
if(cnt>=m) return 1;
}
return 0;
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
cin>>n>>m;re(i,1,n) cin>>a[i],sum[i]=sum[i-1]+a[i];
if(sum[n]<m){
cout<<-1;
return 0;
}
else if(sum[n]==m){
cout<<n;
return 0;
}
sort(a+1,a+1+n,greater<int>());
cout<<half(1,n);
return 0;
}
E. Photo Processing
把序列分段，每段个数大于等于k，最小化每一段的最大值与最小值之差
n方的dp转移方程式是很好写的，如下：
dp[i]=min(dp[i],max(dp[j],a[i]-a[j+1]))
这里dp存的是最小的答案
但是这东西没法用单调队列维护，很鸡屎，我用各种乱七八糟方法都TLE，甚至猜测它是不是一个单峰函数
后来看题解，应该二分最大值与最小值之差
用dp[i]来表示1到i这个位置能不能有合法的分段方法
首先保证最大值最小值之差，然后保证每一段个数，合并的是[last,i]这个区间
最后看看dp[n]是否等于1就完事了
代码：
#include <bits/stdc++.h>
#define int long long
#define sc(a) scanf("%lld",&a)
#define scc(a,b) scanf("%lld %lld",&a,&b)
#define sccc(a,b,c) scanf("%lld %lld %lld",&a,&b,&c)
#define scs(a) scanf("%s",a)
#define schar(a) scanf("%c",&a)
#define pr(a) printf("%lld",a)
#define fo(i,a,b) for(int i=a;i<b;++i)
#define re(i,a,b) for(int i=a;i<=b;++i)
#define rfo(i,a,b) for(int i=a;i>b;--i)
#define rre(i,a,b) for(int i=a;i>=b;--i)
#define prn() printf("\n")
#define prs() printf(" ")
#define mkp make_pair
#define pii pair<int,int>
#define pub(a) push_back(a)
#define pob() pop_back()
#define puf(a) push_front(a)
#define pof() pop_front()
#define fst first
#define snd second
#define frt front()
#define bak back()
#define mem0(a) memset(a,0,sizeof(a))
#define memmx(a) memset(a,0x3f3f,sizeof(a))
#define memmn(a) memset(a,-0x3f3f,sizeof(a))
#define debug
#define db double
#define yyes cout<<"YES"<<endl;
#define nno cout<<"NO"<<endl;
#define all(i,a) for(auto i=a.begin();i!=a.end();++i)
using namespace std;
typedef vector<int> vei;
typedef vector<pii> vep;
typedef map<int,int> mpii;
typedef map<char,int> mpci;
typedef map<string,int> mpsi;
typedef deque<int> deqi;
typedef deque<char> deqc;
typedef priority_queue<int> mxpq;
typedef priority_queue<int,vector<int>,greater<int> > mnpq;
typedef priority_queue<pii> mxpqii;
typedef priority_queue<pii,vector<pii>,greater<pii> > mnpqii;
const int maxn=500005;
const int inf=0x3f3f3f3f3f3f3f3f;
const int MOD=1e9+7;
const db eps=1e-10;
const db pi=3.1415926535;
int qpow(int a,int b){int tmp=a%MOD,ans=1;while(b){if(b&1){ans*=tmp,ans%=MOD;}tmp*=tmp,tmp%=MOD,b>>=1;}return ans;}
int lowbit(int x){return x&-x;}
int max(int a,int b){return a>b?a:b;}
int min(int a,int b){return a<b?a:b;}
int mmax(int a,int b,int c){return max(a,max(b,c));}
int mmin(int a,int b,int c){return min(a,min(b,c));}
void mod(int &a){a+=MOD;a%=MOD;}
bool chk(int now);
int half(int l,int r){while(l<=r){int m=(l+r)/2;if(chk(m))r=m-1;else l=m+1;}return l;}
int ll(int p){return p<<1;}
int rr(int p){return p<<1|1;}
int mm(int l,int r){return (l+r)/2;}
int lg(int x){if(x==0) return 1;return (int)log2(x)+1;}
bool smleql(db a,db b){if(a<b||fabs(a-b)<=eps)return true;return false;}
bool bigeql(db a,db b){if(a>b||fabs(a-b)<=eps)return true;return false;}
bool eql(db a,db b){if(fabs(a-b)<eps) return 1;return 0;}
db len(db a,db b,db c,db d){return sqrt((a-c)*(a-c)+(b-d)*(b-d));}
bool isp(int x){if(x==1)return false;if(x==2)return true;for(int i=2;i*i<=x;++i)if(x%i==0)return false;return true;}
inline int read(){
char ch=getchar();int s=0,w=1;
while(ch<48||ch>57){if(ch==‘-‘)w=-1;ch=getchar();}
while(ch>=48&&ch<=57){s=(s<<1)+(s<<3)+ch-48;ch=getchar();}
return s*w;
}
inline void write(int x){
if(x<0)putchar(‘-‘),x=-x;
if(x>9)write(x/10);
putchar(x%10+48);
}
int gcd(int a, int b){
if(a==0) return b;
if(b==0) return a;
if(!(a&1)&&!(b&1)) return gcd(a>>1,b>>1)<<1;
else if(!(b&1)) return gcd(a,b>>1);
else if(!(a&1)) return gcd(a>>1,b);
else return gcd(abs(a-b),min(a,b));
}
int lcm(int x,int y){return x*y/gcd(x,y);}
int n,k,a[maxn],dp[maxn];
bool chk(int now){
mem0(dp);
dp[0]=1;
int last=1;
re(i,1,n){
while(a[i]-a[last]>now) last++;
while(i-last+1>=k){
if(dp[last-1]){
dp[i]=1;
break;
}
last++;
}
}
return dp[n];
}
signed main(){
ios_base::sync_with_stdio(0);
cin.tie(0),cout.tie(0);
n=read(),k=read();re(i,1,n) a[i]=read();
sort(a+1,a+1+n);
cout<<half(0,1e18);
return 0;
}
原文地址：https://www.cnblogs.com/oneman233/p/11600215.html
时间： 2024-11-09 09:36:08