The 19th Zhejiang Provincial Collegiate Programming Contest
The 19th Zhejiang Provincial Collegiate Programming Contest
今天跑去239和学姐vp 结果发现自己啥也不会QAQ
A. JB Loves Math
只能加上奇数或者减去偶数，且这个数字是固定的
分为三大类：
- \(a==b\)，答案为0；
- \(a > b\),\(a-b\)为奇数，需要两步（先减去\(a-b+1\)，再+1）；为偶数则需要一步；
- \(a < b\),\(b-a\)为偶数，需要一步；为偶数时再分为两种情况讨论，具体见Code:
#include <bits/stdc++.h>
using namespace std;
int main () {
int t;
cin >> t;
while (t --) {
int a, b;
cin >> a >> b;
if (a > b) {
if ((a - b) % 2 == 0)
cout << 1 << endl;
else
cout << 2 << endl;
}
else if (a == b)
cout << 0 << endl;
else {
if ((b - a) % 2)
cout << 1 << endl;
else if ((b - a) / 2 % 2)
cout << 2 << endl;
else
cout << 3 << endl;
}
}
}
//改变奇偶性
//eg.2->6: +3 +3 -2
//答案不会超过3
B. JB Loves Comma
灰常灰常简单的签到题
#include <iostream>
using namespace std;
int main () {
string s;
cin >> s;
int n = s.size ();
if (n < 3)
cout << s << endl;
else {
for (int i = 0; i < s.size (); i ++) {
cout << s[i];
if (s[i] == 'b' && s[i - 1] == 'j' && s[i - 2] == 'c')
cout << ',';
}
cout << endl;
}
}
C. JB Wants to Earn Big Money
题意已经很明确了，序列\({a_n}\)中\(\geq x\)的就统计进ans; 序列\({b_n}\)中\(\leq x\)的就统计进ans
#include <iostream>
using namespace std;
int ans;
int main () {
int n, m, x;
cin >> n >> m >> x;
for (int i = 1; i <= n; i ++) {
int t;
cin >> t;
if (t >= x)
ans ++;
}
for (int i = 1; i <= m; i ++) {
int t;
cin >> t;
if (t <= x)
ans ++;
}
cout << ans << endl;
}
L. Candy Machine
其实我隐约有这个思路了，就是全部扫一遍。。但是我以为太暴力了就没说hhh下次还是要大胆一些！
思路：先排个序，然后按顺序挨个往里面加数，二分找出最大的该区间内严格大于该区间平均数的数的数量，记录一个最大值输出。
stl实现二分查找详细用法mark：
https://zh.cppreference.com/w/cpp/algorithm/upper_bound
https://zh.cppreference.com/w/cpp/algorithm/lower_bound
还有几道题目明天再补
#include <bits/stdc++.h>
#define int long long
using namespace std;
const int N = 1e6 + 5;
int n, a[N], ans;
signed main () {
scanf ("%d", &n);
for (int i = 1; i <= n; i ++)
scanf ("%d", &a[i]);
sort (a + 1, a + n + 1);
int sum = 0;
for (int i = 1; i <= n; i ++) {
sum += a[i];
double aver = 1.0 * sum / i;
int pos = upper_bound (a + 1, a + i + 1, aver) - a;
ans = max (ans, i - pos + 1);
}
printf ("%d\n", ans);
}
//upper_bound是返回首个大于v的迭代器
//lower_bound是返回首个大于等于v的迭代器
//卧槽。。卡cin
M. BpbBppbpBB
本以为是一个很可怕的大模拟，没想到就是解方程orz
就是分别统计黑块和白块（注意不能统计到边上的了）的数目；第一种有黑块146个，白块24个；第二种有黑块100个，白块12个。
*这里统计白块就是看最左上角那个满不满足这样的图形：
####
#..#
#....#
#....#
#..#
看看check函数就明白了：
然后手动解方程，得
\[\begin{cases}
146x+100y=black\\
24x+12y=white -> 2x+y=white'(大白块)
\end{cases}
\,\,\,\,\,\rightarrow
\begin{cases}
x = \frac{100white - black}{54}\\
y = \frac{black - 73white}{27}
\end{cases}\\
\]
#include <bits/stdc++.h>
using namespace std;
const int N = 1005;
int n, m;
char a[N][N];
bool check (int x, int y) {
if (a[x-1][y-1] == '#' && a[x-1][y] == '#' && a[x-1][y+1] == '#' && a[x-1][y+2] == '#' &&
a[x][y-1] == '#' && a[x][y] == '.' && a[x][y+1] == '.' && a[x][y+2] == '#' &&
a[x+1][y-2] == '#' && a[x+1][y-1] == '.' && a[x+1][y] == '.' && a[x+1][y+1] == '.' && a[x+1][y+2] == '.' && a[x+1][y+3] == '#' &&
a[x+2][y-2] == '#' && a[x+2][y-1] == '.' && a[x+2][y] == '.' && a[x+2][y+1] == '.' && a[x+2][y+2] == '.' && a[x+2][y+3] == '#' &&
a[x+3][y-1] == '#' && a[x+3][y] == '.' && a[x+3][y+1] == '.' && a[x+3][y+2] == '#'
)
return true;
return false;
}
int main () {
cin >> n >> m;
int black = 0, white = 0;
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= m; j ++) {
cin >> a[i][j];
if (a[i][j] == '#')
black ++;
}
//统计非边界白块
for (int i = 1; i <= n; i ++)
for (int j = 1; j <= m; j ++) {
//按照左上角来看
if (check (i, j))
white ++;
}
cout << (100*white - black) / 54 << ' ' << (black - 73*white) / 27 << endl;
}
//统计白点和黑点的个数然后解方程
//注意边界也是白块啊
//146x+100y=black
//24x+12y=white -> 2x+y=white'(大白块)
//x = (100*white - black) / 54;
//y = (black - 73*white) / 27;
//笑死。。有人解方程都会解错
