AtCoder Beginner Contest 122

目录

• A - Double Helix
• B - ATCoder
• C - GeT AC
• D - We Like AGC

A - Double Helix

#include<bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;

int main(){
    char c;
    cin >> c;
    if (c == 'A') cout << 'T' << endl;
    if (c == 'T') cout << 'A' << endl;
    if (c == 'C') cout << 'G' << endl;
    if (c == 'G') cout << 'C' << endl;
    return 0;
}

B - ATCoder

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
string s;
int res, num;
int main() {
    cin >> s;
    for (int i = 0; i < s.size(); i++) {
        if (s[i] != 'A' && s[i] != 'T' && s[i] != 'C' && s[i] != 'G') {
            res = max(res, num);
            num = 0;
        } else
            num++;
    }
    res = max(res, num);
    cout << res << endl;
    return 0;
}

C - GeT AC

#include <bits/stdc++.h>

using namespace std;

const int N = 1e6 + 5;
typedef long long LL;
int n, m;
string s;
int sum[N];
int main() {
    cin >> n >> m;
    cin >> s;
    s = " " + s;
    for (int i = 1; i < s.size(); i++) {
        sum[i] += (s[i] == 'C' && s[i-1] == 'A')+sum[i-1];
    }
    while(m--){
        int x, y;
        cin >> x >> y;
        cout << sum[y] - sum[x - 1] -(s[x-1]=='A'&&s[x]=='C')<< endl;
    }
    return 0;
}

D - We Like AGC

给出一个数n，问长度为n的、满足下列条件的字符串有多少个：

字符均由A G C T组成

不包含AGC的子串

交换一次相邻的字符后，也不包含AGC的子串

dp求解，(dp[len][i][j][k])代表长度为len，以i j k为结尾的字符串的数量，然后就很好dp了

#include <bits/stdc++.h>

using namespace std;

const int N = 1e2 + 5;
typedef long long LL;
LL const mod = 1e9 + 7;
LL dp[N][5][5][5];

bool check(int q, int j, int k, int l) {
    // AGCT
    if (q == 0 && j == 1 && k == 2) return 0;
    if (q == 0 && k == 1 && l == 2) return 0;
    if (q == 0 && j == 1 && l == 2) return 0;
    if (j == 0 && k == 1 && l == 2) return 0;
    if (j == 0 && q == 1 && k == 2) return 0;
    if (j == 0 && l == 1 && k == 2) return 0;
    if (q == 0 && k == 1 && j == 2) return 0;
    if (k == 0 && j == 1 && l == 2) return 0;
    return 1;
}
bool check2(int i, int j, int k) {
    // AGCT
    if (i == 0 && j == 1 && k == 2) return 0;
    if (i == 0 && k == 1 && j == 2) return 0;
    if (j == 0 && i == 1 && k == 2) return 0;
    return 1;
}

int main() {
    int n;
    cin >> n;
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            for (int k = 0; k < 4; k++) {
                if (check2(i,j,k))
                dp[3][i][j][k] = 1;
            }
        }
    }

    for (int i = 4; i <= n; i++) {
        for (int j = 0; j < 4; j++) {
            for (int k = 0; k < 4; k++) {
                for (int l = 0; l < 4; l++) {
                    for (int q = 0; q < 4; q++) {
                        if (check(q, j, k, l))
                            dp[i][j][k][l] =
                                (dp[i][j][k][l] + dp[i - 1][q][j][k]) % mod;
                    }
                }
            }
        }
    }
    LL res = 0;
    for (int i = 0; i < 4; i++) {
        for (int j = 0; j < 4; j++) {
            for (int k = 0; k < 4; k++) {
                res = (res + dp[n][i][j][k]) % mod;
            }
        }
    }
    cout << res << endl;
    return 0;
}