Task On The Board CodeForces - 1367D（思维）

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.

思路：对于一个字符串，肯定有最大的字母和次大的字母等等，对于一开始的b数组来说，如果为0的话，那么这个位置一定是最大字母，然后再减去这个位置对于其他位置的贡献，下一个为0的位置一定就是次大的字母，一直这样弄下去直到所有的位置都弄完。
思路很好想，但是实现起来有些困难。
①对于很多个0，怎么计算所有的0的贡献。
②对于一个字母来说，这个字母用完之后，大于等于这个字母的字母个数都要清零。
这是我遇到的两个坑点。
代码如下：

#include<bits/stdc++.h>
#define ll long long
using namespace std;const int maxx=1e2+10;
int vis[maxx],b[maxx];
char s1[maxx];
string s;
int n;inline bool cmp(char a,char b)
{return a>b;
}
int main()
{int t;scanf("%d",&t);while(t--){cin>>s;scanf("%d",&n);for(int i=0;i<26;i++) vis[i]=0;for(int i=1;i<=n;i++) scanf("%d",&b[i]);for(int i=0;i<s.length();i++) vis[s[i]-'a']++;sort(s.begin(),s.end());int cnt=0;while(cnt<n){int num=0;for(int i=1;i<=n;i++) if(b[i]==0) num++;int pos;for(int i=25;i>=0;i--){if(vis[i]>=num){pos=i;break;}}int rr=num;for(int i=pos;i<26;i++) vis[i]=0;num=0;for(int i=1;i<=n;i++){if(b[i]==0){num+=i;s1[i]=(char)('a'+pos);}}int zz=0,z1=0;for(int i=1;i<=n;i++){if(b[i]==0) zz+=i,z1++,b[i]=-1,cnt++;else if(b[i]==-1) continue;else{b[i]-=((i*z1-zz)+(num-zz)-i*(rr-z1));}}//for(int i=1;i<=n;i++) cout<<b[i]<<" ";cout<<endl;}for(int i=1;i<=n;i++) cout<<s1[i];cout<<endl;}return 0;
}

努力加油a啊，(^o)/~

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