題目描述
Polycarp wrote on the board a string s containing only lowercase Latin letters (‘a’-‘z’). This string is known for you and given in the input.
After that, he erased some letters from the string s, and he rewrote the remaining letters in any order. As a result, he got some new string t. You have to find it with some additional information.
Suppose that the string t has length m and the characters are numbered from left to right from 1 to m. You are given a sequence of m integers: b1,b2,…,bm, where bi is the sum of the distances |i−j| from the index i to all such indices j that tj>ti (consider that ‘a’<‘b’<…<‘z’). In other words, to calculate bi, Polycarp finds all such indices j that the index j contains a letter that is later in the alphabet than ti and sums all the values |i−j|.
For example, if t = “abzb”, then:
since t1=‘a’, all other indices contain letters which are later in the alphabet, that is: b1=|1−2|+|1−3|+|1−4|=1+2+3=6;
since t2=‘b’, only the index j=3 contains the letter, which is later in the alphabet, that is: b2=|2−3|=1;
since t3=‘z’, then there are no indexes j such that tj>ti, thus b3=0;
since t4=‘b’, only the index j=3 contains the letter, which is later in the alphabet, that is: b4=|4−3|=1.
Thus, if t = “abzb”, then b=[6,1,0,1].
Given the string s and the array b, find any possible string t for which the following two requirements are fulfilled simultaneously:
t is obtained from s by erasing some letters (possibly zero) and then writing the rest in any order;
the array, constructed from the string t according to the rules above, equals to the array b specified in the input data.
Input
The first line contains an integer q (1≤q≤100) — the number of test cases in the test. Then q test cases follow.
Each test case consists of three lines:
the first line contains string s, which has a length from 1 to 50 and consists of lowercase English letters;
the second line contains positive integer m (1≤m≤|s|), where |s| is the length of the string s, and m is the length of the array b;
the third line contains the integers b1,b2,…,bm (0≤bi≤1225).
It is guaranteed that in each test case an answer exists.
Output
Output q lines: the k-th of them should contain the answer (string t) to the k-th test case. It is guaranteed that an answer to each test case exists. If there are several answers, output any.
Example
input
4
abac
3
2 1 0
abc
1
0
abba
3
1 0 1
ecoosdcefr
10
38 13 24 14 11 5 3 24 17 0
output
aac
b
aba
codeforces
Note
In the first test case, such strings t are suitable: "aac’, “aab”.
In the second test case, such trings t are suitable: “a”, “b”, “c”.
In the third test case, only the string t equals to “aba” is suitable, but the character ‘b’ can be from the second or third position.
題目大意
一段由小寫字母a-z組成的字符串s,你將從中刪除幾個字母並且隨機排序得到字符串t。對於字符串t中的字符t[i]有着與之相對應的b[i]，b[i]爲字符串t中除了t[i]本身,大於t[i]的字符的索引值與i的差的絕對值之和。現在已知字符串s，t的長度和b[i]。求字符串t。
題目分析
- 首先找到b[i]=0的位置，因爲b[i]=0，根據題意，該位置的字母肯定是s中最大的。
再讓其他位置的數都減去abs(i-j),這樣便又能找到新的b[i]=0的位置，那麼該位置的字母便是s中第二大的數。- 以此類推，便能找出最後的答案。
代碼如下
#include <iostream>
#include <cstdio>
#include <cmath>
#include <string>
#include <cstring>
#include <stack>
#include <map>
#include <unordered_map>
#include <queue>
#include <vector>
#include <set>
#include <algorithm>
#include <iomanip>
#define LL long long
using namespace std;
const int N=105;
int a[N],b[N]; //a儲存字符串s中的字母，b爲題目中的b數組
bool st[N]; //記錄該位置是否被用過了
char ans[N]; //答案串
int main()
{
int t;
cin>>t;
while(t--)
{
memset(a,0,sizeof a);
memset(st,0,sizeof st);
memset(ans,0,sizeof ans);
string s;
int n;
cin>>s;
cin>>n;
for(int i=0;i<n;i++)
cin>>b[i];
for(int i=0;i<s.size();i++) //將s中的字符儲存在a中
{
a[s[i]-'a']++;
}
int num=0,k=26; //k記錄當前的最大字符
while(num!=n)
{
int cnt=0; //統計這一輪b[i]==0的個數
for(int i=0;i<n;i++)
if(!st[i]&&b[i]==0) cnt++;
for(int i=k-1;i>=0;i--)
if(a[i]>=cnt) {k=i; break;}
for(int i=0;i<n;i++)
if(!st[i]&&b[i]==0)
{
num++;
st[i]=true;
ans[i]=k+'a';
}
for(int i=0;i<n;i++)
if(b[i]==0&&ans[i]==k+'a')
{
for(int j=0;j<n;j++)
if(b[j]) b[j]-=abs(i-j);
}
}
cout<<ans<<endl;
}
return 0;
}