【58.33%】【codeforces 747B】Mammoth's Genome Decoding

time limit per test1 second
memory limit per test256 megabytes
inputstandard input
outputstandard output
The process of mammoth’s genome decoding in Berland comes to its end!

One of the few remaining tasks is to restore unrecognized nucleotides in a found chain s. Each nucleotide is coded with a capital letter of English alphabet: ‘A’, ‘C’, ‘G’ or ‘T’. Unrecognized nucleotides are coded by a question mark ‘?’. Thus, s is a string consisting of letters ‘A’, ‘C’, ‘G’, ‘T’ and characters ‘?’.

It is known that the number of nucleotides of each of the four types in the decoded genome of mammoth in Berland should be equal.

Your task is to decode the genome and replace each unrecognized nucleotide with one of the four types so that the number of nucleotides of each of the four types becomes equal.

Input
The first line contains the integer n (4 ≤ n ≤ 255) — the length of the genome.

The second line contains the string s of length n — the coded genome. It consists of characters ‘A’, ‘C’, ‘G’, ‘T’ and ‘?’.

Output
If it is possible to decode the genome, print it. If there are multiple answer, print any of them. If it is not possible, print three equals signs in a row: “===” (without quotes).

Examples
input
8
AG?C??CT
output
AGACGTCT
input
4
AGCT
output
AGCT
input
6
????G?

output

input
4
AA??

output

Note
In the first example you can replace the first question mark with the letter ‘A’, the second question mark with the letter ‘G’, the third question mark with the letter ‘T’, then each nucleotide in the genome would be presented twice.

In the second example the genome is already decoded correctly and each nucleotide is exactly once in it.

In the third and the fourth examples it is impossible to decode the genom.

【题目链接】:http://codeforces.com/contest/747/problem/B

【题解】

最后每种碱基肯定都是n/4;
当然如果一开始就大于n/4要输出无解.
如果n不能被4整除;也直接输出无解。
然后枚举每一个问号要换成什么碱基就好;

【完整代码】

#include <bits/stdc++.h>
using namespace std;
#define lson l,m,rt<<1
#define rson m+1,r,rt<<1|1
#define LL long long
#define rep1(i,a,b) for (int i = a;i <= b;i++)
#define rep2(i,a,b) for (int i = a;i >= b;i--)
#define mp make_pair
#define pb push_back
#define fi first
#define se second
#define rei(x) scanf("%d",&x)
#define rel(x) scanf("%I64d",&x)

typedef pair<int,int> pii;
typedef pair<LL,LL> pll;

//const int MAXN = x;
const int dx[9] = {0,1,-1,0,0,-1,-1,1,1};
const int dy[9] = {0,0,0,-1,1,-1,1,-1,1};
const char temp[5] = {'0','A','T','C','G'};
const double pi = acos(-1.0);

int n,q =0,a[5];
string s;

int main()
{
    //freopen("F:\\rush.txt","r",stdin);
    cin >> n;
    cin >> s;
    rep1(i,0,n-1)
        if (s[i]=='?')
            q++;
        else
            if (s[i] == 'A')
                a[1]++;
            else
                if (s[i] == 'T')
                    a[2]++;
                else
                    if (s[i] == 'C')
                        a[3] ++;
                    else
                        if (s[i] =='G')
                            a[4]++;
    if ((n%4)!=0)
    {
        puts("===");
        return 0;
    }
    n/=4;
    rep1(i,1,4)
        if (a[i]>n)
        {
            puts("===");
            return 0;
        }
    int len = s.size();
    rep1(i,0,len-1)
        if (s[i]=='?')
        {
            bool fi = false;
            rep1(j,1,4)
                if (a[j]<n)
                {
                    a[j]++;
                    s[i] = temp[j];
                    fi = true;
                    break;
                }
            if (!fi)
            {
                puts("===");
                return 0;
            }
            q--;
        }
    cout << s<<endl;
    return 0;
}