C. Phone Numbers
time limit per test2 seconds
memory limit per test256 megabytes
inputstandard input
outputstandard 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 pi?<?qi and for all j?<?i it is satisfied that pj?=?qj. 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.
Examples
inputCopy
3 3
abc
output
aca
inputCopy
3 2
abc
output
ac
inputCopy
3 3
ayy
output
yaa
inputCopy
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.
题目大意：给一个长度为n的字符串S，输出一个大于S的字典序的字符串中字典序最小的长度为k的字符串（考试的时候硬是没看懂T.T一直以为输出字典序最小的字符串）
分析：如果k<=n只用从后往前赋值，如果可以找到一个比该位字符字典序大，ans[i]=x,该位前面的直接等于ans[j]=s[j](j=i-1,j>=0,j--)即可，否则，ans[i]=min(s)(字符串s中最小的字母);
如果k>n，i<=n时，ans[i]=s[i];i>n，ans[i]=min(s);
#define debug #include<stdio.h> #include<math.h> #include<cmath> #include<queue> #include<stack> #include<string> #include<cstring> #include<string.h> #include<algorithm> #include<iostream> #include<vector> #include<functional> #include<iomanip> #include<map> #include<set> #define pb push_back using namespace std; typedef long long ll; pair<ll,ll>PLL; pair<int,ll>Pil; const int INF = 0x3f3f3f3f; const double inf=1e8+100; const ll maxn =1e5+100; const int N = 1e4+10; const ll mod=1000007; vector<int>v; char s[maxn],ans[maxn]; int n,k; bool flag; void solve() { int i,j,t=1; //cin>>t; while(t--) { flag=0; vector<int>::iterator it; cin>>n>>k>>s; // cout<<n<<" "<<s<<" "<<k<<endl; for(i=0; i<n; i++) { v.pb(s[i]-‘a‘); } sort(v.begin(),v.end()); v.erase(unique(v.begin(),v.end()),v.end()); for(i=k-1; i>=0; i--) { if(k<=n) { if(flag){ ans[i]=s[i]; continue; } int t=s[i]-‘a‘; it=upper_bound(v.begin(),v.end(),t); if(it==v.end()) { ans[i]=v[0]+‘a‘; // cout<<ans[i]<<" k<=n "; } else { ans[i]=*it+‘a‘; // cout<<ans[i]<<" k<=n "; flag=1; } } else { if(flag) { ans[i]=s[i]; // cout<<ans[i]<<" k>n "; } else { for(;i>=n;i--){ ans[i]=v[0]+‘a‘; // cout<<ans[i]<<" k>n "; } flag=1; i++;//这里不要写落了 } } } // cout<<endl; cout<<ans<<endl; // v.clear(); // memset(ans,‘\0‘,sizeof(ans)); } } int main() { ios_base::sync_with_stdio(false); #ifdef debug freopen("in.txt", "r", stdin); // freopen("out.txt","w",stdout); #endif cin.tie(0); cout.tie(0); solve(); return 0; }