CodeForces - 940C Phone Numbers （字符串，模拟）

And where the are the phone numbers?

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

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 ofabadaba is {a, b, d}.

String p is lexicographically smaller than stringq, if p is a prefix ofq, is not equal to q or there exists i, such thatpi < qi and for allj < i it is satisfied that pj = qj. For example,abc is lexicographically smaller than abcd , abd is lexicographically smaller thanabec, afais 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 ofs and the required length of t.

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

Output
Output the string t conforming to the requirements above.

It’s guaranteed that the answer exists.

Example
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 oft is a subset of letters of s is as follows: aaa, aab, aac, aba, abb, abc, aca, acb, …. Among them, those are lexicographically greater thanabc: aca,acb, …. Out of those the lexicographically smallest isaca.
题意分析：
这道题就是写出比给出字符串字典序大的最小字符串，要求所用字符串的字符来自与给出的原来字符串。
比如说给出
3 3（原字符串的长度，输出字符串的长度）
ayy
输出
yaa

最近刚刚学STL，写的不好。

通过写这道题目，知道了set不能访问指定位置的字符串，只能遍历访问。 知道了vecotr的尾部元素的地址是ite=ve.end()-1

#include
using namespace std;
#define inf 0x3f3f3f3f
#define For(a,b) for(int a=0;a
#define mem(x) memset(x,0,sizeof(x))
#define Debug(x) cout<<"---> "<
#define sf scanf
#define pf printf
int gcd(int a,int b){ return b>0?gcd(b,a%b):a;}
typedef long long ll;
typedef pair<int ,int > P;
//head
#define maxn 300100
int n,k;
char str[maxn];
int A[maxn];
set <int> ss;
set<int>::iterator ite;
vector<int> ve;
vector<int>::iterator it;
int B[maxn],len;
char a;

int main(){
    int pos;
    cin>>n>>k;
    For(i,n) {
        cin>>str[i];
        A[i]=str[i]-'0'-49;
        ss.insert(A[i]);
    }
    for(ite=ss.begin();ite!=ss.end();++ite){
        ve.push_back(*ite);
        B[len++]=*ite;
    }
    it=ve.end()-1;          //访问尾部元素的注意点 
    for(int i=k-1;i>=0;i--){
        if(A[i]==*it){
            A[i]=*ve.begin();
        }
        else{
            int pos=upper_bound(B,B+len,A[i])-B;
            A[i]=B[pos];
            break;
        }
    }

    //print
    if(k<=n){
        for(int i=0;i49+'0';
        cout<else{
        for(int i=0;i49+'0';
            cout<for(int i=0;i0]+49+'0';
            cout<cout<return 0;
}

#include 
#include   
#include   
#include 
#include   
#include   
#include   
#include    
#include   
#include   
#include   
#include 
#include  
#define inf 0x3f3f3f
#define sf scanf
#define pf printf
using namespace std;
typedef long long ll;
typedef pair<int ,int > P;
ll gcd(ll a,ll b) { return b>0?gcd(b,a%b):a;}
ll lcm(ll a,ll b){ return a*b/gcd(a,b);}
//head
#define maxn 20000
using namespace std;
typedef long long ll;
char str;
set<int> s;
int ha[maxn];
int A[maxn],B[maxn];
int n,k;
int main(){
    cin>>n>>k;
    for(int i=0;icin>>str;
        A[i]=str-'a';
        s.insert(A[i]);
    }
    int p=0;
    set<int>::iterator ite;
    for(ite=s.begin();ite!=s.end();++ite){
        B[p++]=*ite;
    }
    p=s.size()-1;

    for(int i=n-1;i>=0;i--){
        if(A[i]==B[p]){
            A[i]=B[0];
        }
        else{
            int pos=upper_bound(B,B+p,A[i])-B;
            A[i]=B[pos];
            break;
        }
    }
    //print
    if(k<=n){
        for(int i=0;i'a';
            cout<else{
        for(int i=0;i'a';
            cout<for(int i=0;i0]+'a';
            cout<cout<return 0;
}