$dp[x][p][pp]$表示第x位,当前已有数字mod 2520(1~9数字的lcm)为p,当前各位数字的lcm为pp
观察到数组太大,考虑压缩,第三维lcm最多只有9个数字,打表发现最多只有48个状态,压掉第三维即可
打表用一个状压然后set维护(广搜也可以)即可
有一个坑点:题目里似乎没有说关于0的事情(即数字里出现0)但是有人在CF上打这个比赛的时候问了出题人,碰到0不要管即可!!!
打表代码:
1 set<int>s; 2 inline void Make(int x){ 3 int ans=1; 4 for(int i=0;i<9;i++){ 5 if(((1<<i)&x)) ans=lcm(ans,i+1); 6 }s.insert(ans); 7 } 8 inline void States_Maker(){ 9 int ans=1; 10 for(int i=0;i<(1<<9);i++)Make(i); 11 while(!s.empty()){ 12 cout<<*s.begin()<<","; 13 s.erase(s.begin()); 14 } 15 }
解题代码:
1 #include<bits/stdc++.h> 2 #define int long long 3 #define writeln(x) write(x),puts("") 4 #define writep(x) write(x),putchar(‘ ‘) 5 using namespace std; 6 inline int read(){ 7 int ans=0,f=1;char chr=getchar(); 8 while(!isdigit(chr)){if(chr==‘-‘) f=-1;chr=getchar();} 9 while(isdigit(chr)){ans=(ans<<3)+(ans<<1)+chr-48;chr=getchar();} 10 return ans*f; 11 }void write(int x){ 12 if(x<0) putchar(‘-‘),x=-x; 13 if(x>9) write(x/10); 14 putchar(x%10+‘0‘); 15 }const int M = 20,State = 50; 16 int f[M][2521][State],T,n,a[M],l,r,Pos[2521],S[State]={0,1,2,3,4,5,6,7,8,9,10,12,14,15,18,20,21,24,28,30,35,36,40,42,45,56,60,63,70,72,84,90,105,120,126,140,168,180,210,252,280,315,360,420,504,630,840,1260,2520}; 17 inline int gcd(int x,int y){return __gcd(x,y);} 18 inline int lcm(int x,int y){if(x==0)return y;if(y==0)return x;return x*y/gcd(x,y);} 19 int dfs(int x,int p,int pp,bool lim){ 20 if(!x)return (p%pp==0)?(1):(0); 21 if(!lim&&p!=-1&&pp!=-1&&f[x][p][Pos[pp]]!=-1)return f[x][p][Pos[pp]]; 22 int ans=0,up=lim?a[x]:9; 23 for(int i=0;i<=up;i++)ans+=dfs(x-1,(p*10+i)%2520,lcm(pp,i),lim&&a[x]==i); 24 if(!lim)f[x][p][Pos[pp]]=ans; 25 return ans; 26 } 27 inline int Solve(int x){ 28 int len=0; 29 while(x)a[++len]=x%10,x/=10; 30 return dfs(len,0,1,1); 31 } 32 signed main(){ 33 memset(f,-1,sizeof(f)); 34 for(int i=1;i<=48;i++)Pos[S[i]]=i; 35 T=read(); 36 while(T--)l=read(),r=read(),writeln(Solve(r)-Solve(l-1)); 37 return 0; 38 }
原文地址:https://www.cnblogs.com/zhenglw/p/11594843.html
时间: 2024-10-21 12:29:03