A、这波啊，这波是.....

这波是签到题，拼手速，哎嘿拿了个一血！

print('roudancongji')

B、李在赣神魔

思维题，旋转90度输出，改一下输出的下标即可。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll 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]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e3 + 7;
char s[N][N];

int main() {
    js;
    int n;
    cin >> n;
    for (int i = 1; i <= n; ++i)
        for (int j = 1; j <= n; ++j)    cin >> s[i][j];
    for (int i = 1; i <= n; ++i) {
        for (int j = 1; j <= n; ++j)
            cout << s[n - j + 1][i];
        cout << '\n';
    }
    return 0;
}

C、电竞希金斯

思维题，考察高中斜率和截距的知识，如果WA的小伙伴）比如我，就是不够仔细。。。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll 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]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;
const double eps = 1e-6;

int main(){
    int a = read(),b = read(),c = read();
    if((a==0 || b == 0) && c == 0){
        puts("non");
        return 0;
    }
    if(a == 0){
        if(b * c > 0) puts("3 4");
        else puts("1 2");
        return 0;
    }
    if(b == 0){
        if( a * c > 0) puts("2 3");
        else puts("1 4");
        return 0;
    }
    if(c == 0){
        if( a* b < 0) puts("1 3");
        else puts("2 4");
        return 0;
    }
    if(a>0&&b>0&&c>0) puts("2 3 4");
    else if( a > 0 && b > 0 && c < 0) puts("1 2 4");
    else if(a > 0 && b < 0 && c > 0) puts("1 2 3");
    else if(a > 0 && b < 0 && c < 0) puts("1 3 4");
    else if(a < 0 && b > 0 && c > 0 ) puts("1 3 4");
    else if( a < 0 && b > 0 && c < 0 ) puts("1 2 3");
    else if( a < 0 && b < 0 && c > 0) puts("1 2 4");
    else puts("1 2 4");

}

F、这题多捞啊

打表！打表！再打表，思维题搞不定，那么就去打表，反正白瞎也是错。甚至我打的表都是只满足条件1，不满足条件2。
最终盲猜偶数1一个输出，奇数2个，因为你会发现前面有1很容易得到n。具体证明……咕咕咕，咱不会吖。

#pragma GCC target("avx,sse2,sse3,sse4,popcnt")
#pragma GCC optimize("O2,O3,Ofast,inline,unroll-all-loops,-ffast-math")
#include <bits/stdc++.h>
using namespace std;
#define js ios::sync_with_stdio(false);cin.tie(0); cout.tie(0)
typedef long long ll; typedef unsigned long long ull; typedef long double ld;
const ll MOD = 1e9 + 7;
inline ll read() { ll s = 0, w = 1; char ch = getchar(); for (; !isdigit(ch); ch = getchar()) if (ch == '-') w = -1; for (; isdigit(ch); ch = getchar())    s = (s << 1) + (s << 3) + (ch ^ 48); return s * w; }
inline void write(ll x) { if (!x) { putchar('0'); return; } char F[40]; ll 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]); }
inline ll gcd(ll x, ll y) { return y ? gcd(y, x % y) : x; }
ll qpow(ll a, ll b) { ll ans = 1;    while (b) { if (b & 1)    ans *= a;        b >>= 1;        a *= a; }    return ans; }    ll qpow(ll a, ll b, ll mod) { ll ans = 1; while (b) { if (b & 1)(ans *= a) %= mod; b >>= 1; (a *= a) %= mod; }return ans % mod; }
inline int lowbit(int x) { return x & (-x); }

const int N = 1e5 + 7;

int main() {
    int n = read();
    for (int i = 1; i < n; ++i)    putchar('1'), putchar(32);
    printf("%d\n", n + 1);
    if (n & 1) {
        for (int i = 1; i < n; ++i)
            printf("2 ");
        printf("2\n");
    }
    return 0;
}