hdu

问题

Let A=∑ni=1ai∗10n−i(1≤ai≤9)A=∑i=1nai∗10n−i(1≤ai≤9)(nn is the number of AA’s digits). We call AA as “beautiful number” if and only if a[i]≥a[i+1]a[i]≥a[i+1] when 1≤i

题解

暴力打表 O(n)

我们跑[1,1e9] 的所有满足这个条件的数。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;int a[1299] ={自己打表};int L,R;int ans;int main(){    //freopen("c:/in.txt","r",stdin);    ios::sync_with_stdio(0);cin.tie(0);    int T;    cin>>T;    while(T--){        ans = 0;        cin>>L>>R;        for(int i = 0;i<1299;i++)        if(a[i]>=L&&a[i]<=R) ans++;        cout<<ans<<endl;    }}

暴力打表源代码

/*暴力求解，写入文件*/#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;int find(int x){    int y = x;    int a[12];    int j = 0;    while(x){        a[j++] = x%10;        x/=10;    }    for(int i=0;i<j-1;i++){        if(a[i]==0||a[i+1]<a[i]||a[i+1]%a[i]!=0)return 0;    }    return y;}int main(){     fstream fFile;     int s = 0;     fFile.open("f:\\1.txt", ios::out);    for(int i =1;i<=1000000000;i++){        int yy = find(i);        if(yy){           fFile <<yy<<',';           s++;        }    }   printf("%d\n",s);} 

dfs

最大也就10个位，暴力dfs每一个位。

#include<iostream>#include<cstdio>#include<cstring>using namespace std;typedef long long LL;int ans;int L,R;void dfs(LL num,int pre){    if(num>R) return ;    if(num>=L) ans++;    for(int i=1;i<=pre;i++){        if(pre%i==0) dfs(num*10+i,i);    }}int main(){    #ifdef DEBUG        freopen("C:\\in.txt", "r", stdin);        //freopen("out.txt", "w", stdout);    #endif    ios::sync_with_stdio(0);cin.tie(0);    int T;    cin>>T;    while(T--){        ans=0;        cin>>L>>R;        for(int i=1;i<=9;i++){            dfs(i,i);        }        cout<<ans<<endl;    }}

数位dp

在dfs上加记忆化搜索

#include<iostream>#include<cstdio>#include<cstring>#include<cmath>using namespace std;typedef long long LL;#define ms(a, b) memset(a, b, sizeof(a))int L,R;int ans;int a[15];int dp[25][10];int dfs(int pos,bool lead,bool limit,int pre){    if(pos == 0) return 1;    if(!limit && !lead && dp[pos][pre]!=-1) return dp[pos][pre];    int up=limit?a[pos]:9;    LL ans=0;    for(int i=0;i<=up;i++){        if(lead||pre>=i&&i!=0&&pre%i==0){            ans+=dfs(pos-1,lead&&i==0,limit&&i==a[pos],i);        }    }    if(!limit&&!lead) dp[pos][pre] = ans;    return ans;}int solve(int x){    int len = 0;    while(x){        a[++len] = x%10;        x/=10;    }    return dfs(len, true, true, 0);}int main(){    //freopen("c:/in.txt","r",stdin);    ios::sync_with_stdio(0);cin.tie(0);    int t;    cin>>t;    ms(dp,-1);    while(t--){        cin>>L>>R;        cout<<solve(R)-solve(L-1)<<endl;    }    return 0;}