AtCoder Beginner Contest 122 解题报告

ACM竞赛代码解析

最新推荐文章于 2025-05-11 23:36:53 发布

Angela㐅cc

最新推荐文章于 2025-05-11 23:36:53 发布

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原文链接：http://www.cnblogs.com/henry-1202/p/10590327.html

本文解析了四道ACM竞赛题目(A-DoubleHelix, B-ATCoder, C-GeTAC, D-WeLikeAGC)的代码实现，涵盖字符串处理、快速输入输出技巧、动态规划等算法知识。

~~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);
}

转载于:https://www.cnblogs.com/henry-1202/p/10590327.html