A. Dislike of Threes

简单的水题,预处理即可

AC_CODE

#include <bits/stdc++.h>

using namespace std;

template < typename T >
inline void read(T &x)
{
    x = 0; bool f = 0; char ch = getchar();
    while(!isdigit(ch)){f ^= !(ch ^ 45);ch=getchar();}
    while(isdigit(ch)) x= (x<<1)+(x<<3)+(ch&15),ch=getchar();
    x = f ? -x : x;
}
const int N = 1e5 + 10;
int a[N];
void solve() {
	int n; read(n);
	printf("%d\n", a[n]);
}

signed main()
{
	int p = 1;
	for(int i = 1; p < 1110; i ++ ) {
		if(i % 3 == 0 || i % 10 == 3) continue;
		a[p ++ ] = i;
	}
    int T = 1;cin >> T;
    while(T -- ) solve();
    return 0;
}

B. Who's Opposite?

sb题
找到这个环的中间位置,然后判断三个数字是否在环外即可

AC_CODE

#include <bits/stdc++.h>

using namespace std;

template < typename T >
inline void read(T &x)
{
    x = 0; bool f = 0; char ch = getchar();
    while(!isdigit(ch)){f ^= !(ch ^ 45);ch=getchar();}
    while(isdigit(ch)) x= (x<<1)+(x<<3)+(ch&15),ch=getchar();
    x = f ? -x : x;
}
 
void solve() {
	int a, b, c;
	read(a); read(b); read(c);
	if(a > b) swap(a, b);
	int res = b - a;
	int len = res * 2;
	if(b > len || c > len) {
		puts("-1");
		return;
	}
	int ans = c + res;
	if(ans > 2 * res) ans %= (2 * res);
	
	printf("%d\n",ans);
}
 
signed main()
{
    int T = 1;cin >> T;
    while(T -- ) solve();
    return 0;
}

C - Infinity Table

预处理出所有的平方数

• 首先判断这个数字是在哪一行or哪一列 (开方向上取整即可) 假设这个数字是idx
• 其次判断这个数字是在列还是行
• 如果是列 则输出 n - \(idx^2\) 如果是行则输出 \((idx+1)^2\) - n + 1

#include <bits/stdc++.h>

using namespace std;

template < typename T >
inline void read(T &x)
{
    x = 0; bool f = 0; char ch = getchar();
    while(!isdigit(ch)){f ^= !(ch ^ 45);ch=getchar();}
    while(isdigit(ch)) x= (x<<1)+(x<<3)+(ch&15),ch=getchar();
    x = f ? -x : x;
}
const int N = 1e5 + 10;
int a[N];
int p = 1;
void solve() {
	int n; read(n);
	int idx = lower_bound(a + 1, a + 1 + p, n) - a;
	int res = n - a[idx - 1];
	if(res <= idx) {
		printf("%d %d\n", res, idx);
		return;
	}
	res = a[idx] - n;
	printf("%d %d\n", idx, res + 1);
	
}

signed main()
{

	for(int i = 1; i <= INF / i; i ++ )	 {
		a[p ++] = i * i;
	}

    int T = 1;cin >> T;
    while(T -- ) solve();
    return 0;
}

D - Make a Power of Two

预处理出所有的 \(2^n\)
暴力枚举从原来的数字操作到 \(2^n\) 所需的最小操作次数 取最小值
最小操作次数判断时候,即是判断重复子序列长度

AC_CODE

//#pragma GCC optimize (2)
//#pragma G++ optimize (2)
#include <bits/stdc++.h>

using namespace std;

template < typename T >
inline void read(T &x)
{
    x = 0; bool f = 0; char ch = getchar();
    while(!isdigit(ch)){f ^= !(ch ^ 45);ch=getchar();}
    while(isdigit(ch)) x= (x<<1)+(x<<3)+(ch&15),ch=getchar();
    x = f ? -x : x;
}
 
string a[63];

void solve() {
    string x; cin >> x;
    int ans = INF;
    int p = x.size();
    for(int i = 0; i < 63; i ++ ) {
        int len = a[i].size(), tt = 0;
        for(int j = 0; j < p; j ++ ) {
            if(tt < len && x[j] == a[i][tt]) tt ++; 
        }
        int r = len - tt + p - tt;
        ans = min(ans, r);
        if(ans > r) {
            cout << i << endl;
            ans = r;
        }

    }
    printf("%d\n", ans);
}
 
signed main() 
{
    for(int i = 0; i < 63; i ++ ) {
        LL p = (1LL << i);
        a[i] = to_string(p);
    }
    int T = 1;cin >> T;
    while(T -- ) solve();
    return 0;
}