手速选手成功混进rated only里面的前30名，但是总排名就到110+了...

A - Double Helix

#include <bits/stdc++.h>

#define ll long long

#define inf 0x3f3f3f3f

#define il inline

namespace io {

#define in(a) a = read()

#define out(a) write(a)

#define outn(a) out(a), putchar('\n')

#define I_int ll

inline I_int read() {

    I_int x = 0, f = 1;

    char c = getchar();

    while (c < '0' || c > '9') {

        if (c == '-') f = -1;

        c = getchar();

    }

    while (c >= '0' && c <= '9') {

        x = x * 10 + c - '0';

        c = getchar();

    }

    return x * f;

}

char F[200];

inline void write(I_int x) {

    if (x == 0) return (void) (putchar('0'));

    I_int tmp = x > 0 ? x : -x;

    if (x < 0) putchar('-');

    int cnt = 0;

    while (tmp > 0) {

        F[cnt++] = tmp % 10 + '0';

        tmp /= 10;

    }

    while (cnt > 0) putchar(F[--cnt]);

}

#undef I_int

}

using namespace io;

using namespace std;

#define N 100010

char s[N];

int main() {

	 scanf("%s",s);

	 if(s[0] == 'A') putchar('T');

	 if(s[0] == 'T') putchar('A');

	 if(s[0] == 'G') putchar('C');

	 if(s[0] == 'C') putchar('G');

}

B - ATCoder

#include <bits/stdc++.h>

#define ll long long

#define inf 0x3f3f3f3f

#define il inline

namespace io {

#define in(a) a = read()

#define out(a) write(a)

#define outn(a) out(a), putchar('\n')

#define I_int ll

inline I_int read() {

    I_int x = 0, f = 1;

    char c = getchar();

    while (c < '0' || c > '9') {

        if (c == '-') f = -1;

        c = getchar();

    }

    while (c >= '0' && c <= '9') {

        x = x * 10 + c - '0';

        c = getchar();

    }

    return x * f;

}

char F[200];

inline void write(I_int x) {

    if (x == 0) return (void) (putchar('0'));

    I_int tmp = x > 0 ? x : -x;

    if (x < 0) putchar('-');

    int cnt = 0;

    while (tmp > 0) {

        F[cnt++] = tmp % 10 + '0';

        tmp /= 10;

    }

    while (cnt > 0) putchar(F[--cnt]);

}

#undef I_int

}

using namespace io;

using namespace std;

#define N 100010

char s[N];

bool check(char c) {

	if(c == 'A') return 1;

	if(c == 'G') return 1;

	if(c == 'T') return 1;

	if(c == 'C') return 1;

	return 0;

}

int main() {

	 scanf("%s",s); int cnt = 0, ans = 0, n = strlen(s);

	 for(int i = 0; i < n; ++i) {

	 	if(check(s[i])) ++cnt;

	 	else ans = max(ans, cnt), cnt = 0;

	 }

	 ans = max(ans, cnt);

	 outn(ans);

}

C - GeT AC

这个C怎么比B还简单...

#include <bits/stdc++.h>

#define ll long long

#define inf 0x3f3f3f3f

#define il inline

namespace io {

#define in(a) a = read()

#define out(a) write(a)

#define outn(a) out(a), putchar('\n')

#define I_int ll

inline I_int read() {

    I_int x = 0, f = 1;

    char c = getchar();

    while (c < '0' || c > '9') {

        if (c == '-') f = -1;

        c = getchar();

    }

    while (c >= '0' && c <= '9') {

        x = x * 10 + c - '0';

        c = getchar();

    }

    return x * f;

}

char F[200];

inline void write(I_int x) {

    if (x == 0) return (void) (putchar('0'));

    I_int tmp = x > 0 ? x : -x;

    if (x < 0) putchar('-');

    int cnt = 0;

    while (tmp > 0) {

        F[cnt++] = tmp % 10 + '0';

        tmp /= 10;

    }

    while (cnt > 0) putchar(F[--cnt]);

}

#undef I_int

}

using namespace io;

using namespace std;

#define N 100010

int n = read(), q = read();

int sum[N];

char s[N];

int main() {

	 scanf("%s",s+1);

	 for(int i = 1; i <= n; ++i) {

	 	sum[i] = sum[i - 1];

	 	if(s[i]=='C'&&s[i-1]=='A') {

	 		++sum[i];

		 }

	 }

	 while(q--){

	 	int l = read(), r = read();

	 	outn(sum[r]-sum[l]);

	 }

}

D - We Like AGC

全场唯一有点意思的题了...直接考虑太麻烦了，dp一下。

\(f[i][j][a][b]\)表示位置i，i字符j,i-1字符a,i-2字符b。

利用map来简化判断过程： mp[1]='A'; mp[2]='C';mp[3]='G';mp[4]='T'

利用string重载了加法这个特性，也可以简化判断过程。

然后思考一下就会发现不合法的判定情况就6种（虽然我想+调了要半小时...），如下（a,b,c,d的位置是递增的）

map<int,string>mp;

bool check(int a,int b,int c, int d) {

	if(mp[a]+mp[c]+mp[b]=="AGC") return 0;

	if(mp[a]+mp[b]+mp[d]=="AGC") return 0;

	if(mp[a]+mp[c]+mp[d]=="AGC") return 0;

	if(mp[b]+mp[c]+mp[d]=="AGC") return 0;

	if(mp[b]+mp[d]+mp[c]=="AGC") return 0;

	if(mp[c]+mp[b]+mp[d]=="AGC") return 0;

	return 1;

}

所以转移的时候再多枚举一个第i-3位的字符，就可以\(O(5^4n)\)解决这题了。注意要特判\(n\leq 2\)的情况。

#include <bits/stdc++.h>

#define ll long long

#define inf 0x3f3f3f3f

#define il inline

namespace io {

#define in(a) a = read()

#define out(a) write(a)

#define outn(a) out(a), putchar('\n')

#define I_int ll

inline I_int read() {

    I_int x = 0, f = 1;

    char c = getchar();

    while (c < '0' || c > '9') {

        if (c == '-') f = -1;

        c = getchar();

    }

    while (c >= '0' && c <= '9') {

        x = x * 10 + c - '0';

        c = getchar();

    }

    return x * f;

}

char F[200];

inline void write(I_int x) {

    if (x == 0) return (void) (putchar('0'));

    I_int tmp = x > 0 ? x : -x;

    if (x < 0) putchar('-');

    int cnt = 0;

    while (tmp > 0) {

        F[cnt++] = tmp % 10 + '0';

        tmp /= 10;

    }

    while (cnt > 0) putchar(F[--cnt]);

}

#undef I_int

}

using namespace io;

using namespace std;

#define N 100010

const ll mod = 1e9 + 7;

int n = read();

ll f[N][5][5][5];

/*

f[i][j][a][b]位置i，字符j,i-1字符a,i-2字符b

*/

map<int,string>mp;

bool check(int a,int b,int c, int d) {

	if(mp[a]+mp[c]+mp[b]=="AGC") return 0;

	if(mp[a]+mp[b]+mp[d]=="AGC") return 0;

	if(mp[a]+mp[c]+mp[d]=="AGC") return 0;

	if(mp[b]+mp[c]+mp[d]=="AGC") return 0;

	if(mp[b]+mp[d]+mp[c]=="AGC") return 0;

	if(mp[c]+mp[b]+mp[d]=="AGC") return 0;

	return 1;

}

int main() {

	if(n<=2) return outn(pow(4,n)), 0;

	mp[0]='Q'; mp[1]='A'; mp[2]='C';mp[3]='G';mp[4]='T';

	f[0][0][0][0]=1;

	 for(int i = 1; i <= n; ++i)

	 	for(int j = 1; j <= 4; ++j)

	 		for(int a = 0; a <= 4; ++a)

	 			for(int b = 0; b <= 4; ++b)

	 				for(int l = 0; l <= 4; ++l)

	 					if(check(l,b,a,j)) (f[i][j][a][b] += f[i-1][a][b][l]) %= mod;

	 ll ans = 0;

	 for(int i = 1; i <= 4; ++i) {

	 	for(int j = 1; j <= 4; ++j) {

	 		for(int k = 1; k <= 4; ++k) {

	 			(ans+=f[n][i][j][k])%=mod;

			 }

		 }

	 }

	 outn(ans);

}