非常牛逼的一道路径压缩例题,转移方程非常好想,在数轴上dp即可,可以摸到30分
方程:dp[i]=min{dp[j]+flag[i]}(s<=j<=t)
考虑怎么处理这个最大1e9的巨型数轴,由于s和t的范围很小,所以考虑这个定理:
两个互质的数最大不能表示的数是xy-x-y
详细参见:逆元和威尔逊定理
对于这个连续的区间,它们最大不能表示的数一定小于它们的最小公倍数,所以所有距离大于最小公倍数的点都可以被压缩
这样路径就变得可以处理了,不放心的话还可以把压缩的阈值设定到3000,保证不会出锅
记得最后还要修改一下整个数轴的长度,方便dp结束之后输出答案
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=1000005; 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 L,s,t,n,p[maxn]; int d[maxn],flag[maxn]; int dp[maxn]; signed main(){ ios_base::sync_with_stdio(0); cin.tie(0),cout.tie(0); cin>>L>>s>>t>>n; re(i,1,n) cin>>p[i]; sort(p+1,p+1+n); if(s==t){ int ans=0; re(i,1,n) ans+=(p[i]%s==0); cout<<ans; return 0; } int D=3000; re(i,1,n) d[i]=(p[i]-p[i-1])%D; re(i,1,n){ p[i]=p[i-1]+d[i]; flag[p[i]]=1; } memmx(dp); dp[0]=0; L=p[n]; re(i,1,L+t){ re(j,s,t){ if(i-j>=0) dp[i]=min(dp[i],dp[i-j]); dp[i]+=flag[i]; } } cout<<dp[L+t]; return 0; }