CodeForces - 940C + CodeForces - 932B (两道比较好的模拟题)

940C链接：http://codeforces.com/problemset/problem/940/C

C. Phone Numbers

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

And where the are the phone numbers?

You are given a string s consisting of lowercase English letters and an integer k. Find the lexicographically smallest string t of length k, such that its set of letters is a subset of the set of letters of s and s is lexicographically smaller than t.

It's guaranteed that the answer exists.

Note that the set of letters is a set, not a multiset. For example, the set of letters of abadaba is {a, b, d}.

String p is lexicographically smaller than string q, if p is a prefix of q, is not equal to q or there exists i, such that p_i < q_i and for all j < i it is satisfied that p_j = q_j. For example, abc is lexicographically smaller than abcd , abd is lexicographically smaller than abec, afa is not lexicographically smaller than ab and a is not lexicographically smaller than a.

Input

The first line of input contains two space separated integers n and k (1 ≤ n, k ≤ 100 000) — the length of s and the required length of t.

The second line of input contains the string s consisting of n lowercase English letters.

Output

Output the string t conforming to the requirements above.

It's guaranteed that the answer exists.

input

3 3
abc

output

aca

input

3 2
abc

output

ac

input

3 3
ayy

output

yaa

input

2 3
ba

output

baa

Note

In the first example the list of strings t of length 3, such that the set of letters of t is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, .... Among them, those are lexicographically greater than abc: aca, acb, .... Out of those the lexicographically smallest is aca.

题意就是给你m长的字符串a，输出在给定字符基础上，输出长度为n且字典序恰好比a大的那个字符串。

似乎没坑，直接模拟！

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const int maxn=1e5+;
char s[maxn];
bool vis[];
int main()
{
    int m,n;
    while(cin>>m>>n)
    {
        cin>>s;
        memset(vis,false,sizeof(vis));
        for(int i=; i<m; i++)    vis[s[i]-'a']=true;
        if(m<n)
        {
            for(int i=m; i<n; i++)
            {
                int f=;
                for(int j=; j<=&&f; j++)
                {
                    if(vis[j])
                        f=,s[i]=j+'a';
                }
            }
        }
        else
        {
            int f=;
            for(int i=n-; i>=&&!f; i--)
            {
                int pos=s[i]-'a';
                for(int j=pos+; j<=; j++)
                {
                    if(vis[j])
                    {
                        f=;
                        s[i]=j+'a';
                        break;
                    }
                }
                if(!f)
                {
                    int ff=;
                    for(int j=; j<=&&ff; j++)
                    {
                        if(vis[j])
                            ff=,s[i]=j+'a';
                    }
                }
            }
        }
        for(int i=;i<n;i++)    cout<<s[i];
        cout<<endl;
    }
    return ;
}

932B链接：http://codeforces.com/problemset/problem/932/B

B. Recursive Queries

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Let us define two functions f and g on positive integer numbers.

You need to process Q queries. In each query, you will be given three integers l, r and k. You need to print the number of integers x between l and r inclusive, such that g(x) = k.

Input

The first line of the input contains an integer Q (1 ≤ Q ≤ 2 × 10⁵) representing the number of queries.

Q lines follow, each of which contains 3 integers l, r and k (1 ≤ l ≤ r ≤ 10⁶, 1 ≤ k ≤ 9).

Output

For each query, print a single line containing the answer for that query.

Examples

input

Copy

4
22 73 9
45 64 6
47 55 7
2 62 4

output

Copy

1
4
0
8

input

Copy

4
82 94 6
56 67 4
28 59 9
39 74 4

output

Copy

3
1
1
5

Note

In the first example:

• g(33) = 9 as g(33) = g(3 × 3) = g(9) = 9
• g(47) = g(48) = g(60) = g(61) = 6
• There are no such integers between 47 and 55.
• g(4) = g(14) = g(22) = g(27) = g(39) = g(40) = g(41) = g(58) = 4

g(x)函数满足以下性质：  1. g(x)=x  x<=9

　　　　　　　　　　　2.g(x)=f(x),  x>9

其中，f(x)函数表示的是，将x数位分离，非零的各个位上的数字相乘.

解题思路：

　　第一反应求g(x)时候，记忆化+打表。

　 然后看了下查询区间和查询次数的数据范围，打表时候还是把每个数对应的1--9的个数都求出来，每次查询时候直接一次O(1)的访问就差不多了。

　

1A代码：

#include<cstdio>
#include<iostream>
#include <cstring>
#include <algorithm>
#include <queue>
#include<cmath>
#define ll long long
using namespace std;
const int maxn=1e6+;
int a[maxn][];
int num[maxn];
int get(int x)
{
    if(num[x]!=-) return  num[x];
    else if(x<=)   return x;
    else
    {
        int ans=;
        while(x)
        {
            if(x%!=) ans*=(x%);
            x/=;
        }
        return get(ans);
    }
}
void pre()
{
    memset(num,-,sizeof(num));
    for(int i=;i<=maxn;i++)
        num[i]=get(i);
    //for(int i=20;i<=30;i++) printf("%d  ",num[i]);
    for(int i=;i<=;i++)   a[][i]=;
    a[][]=;
    for(int i=;i<=maxn;i++)
    {
        for(int j=;j<=;j++)
        {
            a[i][j]=a[i-][j];
        }
        int cnt=num[i];
        a[i][cnt]++;
    }
    return ;
}
int main()
{
    int n,x,l,r;
    pre();
    while(~scanf("%d",&n))
    {
        while(n--)
        {
            scanf("%d%d%d",&l,&r,&x);
            if(num[l]==x)
                cout<<a[r][x]-a[l][x]+<<endl;
            else
                cout<<a[r][x]-a[l][x]<<endl;
        }
    }
    return ;
}