哈尔滨华德学院-新生编程挑战赛
哈尔滨华德学院-新生编程挑战赛
A-签到_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
签到
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n;
cin >> n;
map<int,int> mp;
for(int i = 0;i < n;i ++){
int x;
cin >> x;
if(mp.count(x)) continue;
cout << x << ' ';
mp[x] ++;
}
return 0;
}
B-百分之x的信心_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
double n;
cin >> n;
double ans = round(n * 100.0);
int x = ans;
ans = x;
cout << ans << '%';
return 0;
}
C-幻方_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
i64 n;
cin >> n;
i64 x = (n * n + 1) / 2;
cout << x * n << ' ' << x << '\n';
return 0;
}
D-不是幻方_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
行和列逆着输入,就能用\(sort\)对每一列排序,输出也是行列交换输出
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
i64 n;
cin >> n;
vector g(n,vector<int>(n));
for(int i = 0;i < n;i ++)
for(int j = 0;j < n;j ++)
cin >> g[j][i];
for(auto &i : g){
sort(i.begin(),i.end());
}
for(int i = 0;i < n;i ++)
for(int j = 0;j < n;j ++)
cout << g[j][i] << " \n"[j == n - 1];
return 0;
}
E-要长脑子辣！！！_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
\[\begin{aligned}
x &= \frac{\sum_{i=0}^n(a_i\cdot b_j)}{(\sum_{i=0}^na_i)^x+(\sum_{i=0}^n b_i)^y}
\end{aligned}
\]
数据范围是\(1\leq n,a_i,b_i\le 100,1\le x,y\le10\)
如果直接暴力的话可能会产生\(1e40\)的的大数字,而这在\(C++\)中甚至\(int128\)也不行,不知道python
假如用\(a\)表示\(\sum_{i=0}^na_i\),用\(b\)表示\(\sum_{i=0}^nb_i\),用\(c\)表示\(\sum_{i=0}^n(a_i\cdot b_i)\),则可以简化成:
\[\begin{aligned}
x&=\frac{c}{a^x+b^y}\\
&=\frac{1}{\frac{a^x+b^y}{c}}\\
&=\frac{1}{\frac{a^x}{c}+\frac{b^y}{c}}
\end{aligned}
\]
这个时候\(a,b\)还是可能超出范围,那我们可以将\(a,b,c\)进行质因数分解,即:
\[\begin{aligned}
a &= p_0^{a_1}\cdot p_1^{a_2}\cdot p_2^{a_3}\dots\\
b &= p_0^{b_1}\cdot p_1^{b_2}\cdot p_2^{b_3}\dots\\
c &= p_0^{c_1}\cdot p_1^{c_2}\cdot p_2^{c_3}\dots
\end{aligned}
\]
其中\(p_i\)表示质因子,\(a_i/b_i/c_i\)代表次方,然后就是用质因子的次方去进行一个约分,,最后把约分完的因子乘起来就是约分后的值了,假设\(\frac{a^x}{c}\)约分后为\(\frac{xa}{xc}\),\(\frac{b^y}{c}\)约分后为\(\frac{xb}{cc}\),
则有:
\[\begin{aligned}
x&=\frac{1}{\frac{xa}{xc}+\frac{xb}{cc}}\\
&=\frac{xc\cdot cc}{xa\cdot cc + xb\cdot xc}
\end{aligned}
\]
这个时候分子分母都不会超过\(longlong\)范围,可以直接用\(gcd\)函数约分了.
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
long long ksm(long long a, long long b) {
long long res = 1;
while (b) {
if (b & 1)res = res * a;
b >>= 1;
a = a * a;
}
return res;
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
i64 n, x, y;
cin >> n >> x >> y;
i64 a = 0, b = 0, c = 0;
vector<int> sa(n), sb(n);
for (int i = 0, x; i < n; i ++) {
cin >> x;
sa[i] = x;
a += x;
}
for (int i = 0, x; i < n; i ++) {
cin >> x;
sb[i] = x;
b += x;
c += sa[i] * sb[i];
}
vector<PII> A, B, C;
for (int i = 2; i <= a / i; i ++) {
if (a % i == 0) {
i64 res = 0;
while (a % i == 0)
a /= i, res ++;
A.emplace_back(i, res * x);
}
}
if (a > 1) A.emplace_back(a, x);
for (int i = 2; i <= b / i; i ++) {
if (b % i == 0) {
i64 res = 0;
while (b % i == 0)
b /= i, res ++;
B.emplace_back(i, res * y);
}
}
if (b > 1) B.emplace_back(b, y);
for (int i = 2; i <= c / i; i ++) {
if (c % i == 0) {
i64 res = 0;
while (c % i == 0)
c /= i, res ++;
C.emplace_back(i, res);
}
}
if (c > 1) C.emplace_back(c, 1);
i64 xa = 1, xc = 1;
auto CC = C;
int i = 0, j = 0;
for (; i < A.size() && j < C.size();) {
if (A[i].first == C[j].first) {
i64 k = A[i].second;
A[i].second = max(0ll, k - C[j].second);
C[j].second = max(0ll, C[j].second - k);
xa *= ksm(A[i].first, A[i].second), i ++;
xc *= ksm(C[j].first, C[j].second), j ++;
} else if (A[i].first < C[j].first) {
xa *= ksm(A[i].first, A[i].second), i ++;
} else
xc *= ksm(C[j].first, C[j].second), j ++;
}
while (i < A.size()) {
xa *= ksm(A[i].first, A[i].second), i ++;
}
while (j < C.size()) {
xc *= ksm(C[j].first, C[j].second), j ++;
}
i64 xb = 1, cc = 1;
for (i = 0, j = 0; i < B.size() && j < CC.size();) {
if (B[i].first == CC[j].first) {
i64 k = B[i].second;
B[i].second = max(0ll, k - CC[j].second);
CC[j].second = max(0ll, CC[j].second - k);
xb *= ksm(B[i].first, B[i].second), i ++;
cc *= ksm(CC[j].first, CC[j].second), j ++;
} else if (A[i].first < C[j].first) {
xb *= ksm(B[i].first, B[i].second), i ++;
} else
cc *= ksm(CC[j].first, CC[j].second), j ++;
}
while (i < B.size()) {
xb *= ksm(B[i].first, B[i].second), i ++;
}
while (j < CC.size()) {
cc *= ksm(CC[j].first, CC[j].second), j ++;
}
i64 fenzi = xc * cc, fenmu = xa * cc + xb * xc;
i64 g = gcd(fenzi, fenmu);
fenzi /= g, fenmu /= g;
if (fenmu == 1)
cout << fenzi << '\n';
else
cout << fenzi << '/' << fenmu << '\n';
return 0;
}
F-Huadeyyds_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
数据小,直接暴力
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string s , tar = "Huade";
cin >> s;
i64 ans = 0;
string res = "";
for (int i = 0; i < s.size(); i ++) {
if (s.substr(i, 5) == tar) {
ans ++ ;
res += "Huadeyyds";
i += 4;
} else
res += s[i];
}
if (ans)
cout << ans << '\n' << res << '\n';
else
cout << '0';
return 0;
}
G题赛时放错题了来着
H-神奇"?"_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
暴力
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
string s,st;
cin >> s >> st;
int ans = 0;
for(int i = 0;i < s.size();i ++){
string res = s.substr(i,st.size());
int l = 0, r = 0;
while(l < res.size() && (res[l] == st[r] || res[l] == '?')){
l ++, r ++;
}
if(l == st.size()) ans ++;
}
cout << ans << '\n';
return 0;
}
I-Crazy 小飞象！_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
\(bfs\)
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, m, x, y;
cin >> n >> m >> x >> y;
int u[] = {2,2,-2,-2} ,v[] = {2,-2,2,-2};
vector g(n + 1,vector<int>(m + 1,-1));
g[x][y] = 0;
queue<PII> Q;
Q.push({x,y});
vector<bitset<410>> vis(410);
vis[x][y] = 1;
while(Q.size()){
auto [dx,dy] = Q.front();
Q.pop();
for(int i = 0;i < 4;i ++){
int xd = dx + u[i];
int yd = dy + v[i];
if(xd > 0 && xd <= n && yd > 0 && yd <= m && !vis[xd][yd]){
g[xd][yd] = g[dx][dy] + 1;
vis[xd][yd] = 1;
Q.push({xd,yd});
}
}
}
for(int i = 1;i <= n;i ++)
for(int j = 1;j <= m;j ++)
cout << g[i][j] << " \n"[j == m];
return 0;
}
J-幸存者_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
约瑟夫问题变种,就是把第\(k+1\)个循环输出即可
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, k;
cin >> n >> k;
queue<int> Q;
for (int x, i = 1; i <= n; i ++) {
cin >> x;
Q.push(x);
}
Q.push(Q.front());
Q.pop();
int now = 1;
while (Q.size() > 1) {
if (now % (k + 1) == 0) {
cout << Q.front() << ' ';
} else
Q.push(Q.front());
now ++;
Q.pop();
}
return 0;
}
K-喝“水题”_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
using i64 = long long;
typedef pair<i64, i64> PII;
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
double R, r, H, h;
cin >> R >> r >> H >> h;
double ans = 2.0 * 3.14 * R * R * R + 3.14 * (H * R * R - h * r * r);
printf("%.2lf\n", ans / 3.0);
return 0;
}
L-小成背单词_哈尔滨华德学院-新生编程挑战赛（同步赛） (nowcoder.com)
线段树板子题
#include <bits/stdc++.h>
#define debug(a) cout<<#a<<"="<<a<<'\n';
using namespace std;
#define MAXN 100010
#define INF 0x3fffffff
string A[MAXN]; //操作的序列,记得为（1...n）非（0...n）
struct node
{
int left;
int right;
string max; //维护最大值
} Tree[MAXN << 2]; //存储线段树
void maintain(int root) //向上调整，使得让线段树维护区间最小值最大值区间和
{
int LC = root << 1; //此根的左孩子
int RC = (root << 1) + 1; //此根的右孩子
Tree[root].max = max(Tree[LC].max, Tree[RC].max); //根的最大值
}
void Build(int root, int start, int end) //构建线段树
{ //初始化时传入Build(1,1,n);
Tree[root].left = start; //建区间大小
Tree[root].right = end;
if (start == end) //当到达叶子节点时
{
Tree[root].max = A[start];
return;
}
int mid = (start + end) >> 1; //中间分开
Build(root << 1, start, mid); //对左孩子建树，左边孩子的编号为root*2
Build((root << 1) + 1, mid + 1, end); //对右边孩子建树
maintain(root);
}
void update(int root, int pos, string value) //更新点的值
{
if (Tree[root].left == Tree[root].right && Tree[root].left == pos) //更新叶子节点的值
{
Tree[root].max = value;
return;
}
int mid = (Tree[root].left + Tree[root].right) >> 1; //中间分开成两个区间
if (pos <= mid) //更新的值在左孩子
update(root << 1, pos, value); //更新左孩子
else
update((root << 1) + 1, pos, value); //更新的值在右孩子
maintain(root); //叶子节点更新完成后，会回溯到他的父节点，这样一直往上更新到根节点，维护线段树性质
}
string RmaxQ(int root, int start, int end) //查询区间最大值
{
if (start == Tree[root].left && Tree[root].right == end)
{
return Tree[root].max;
}
int mid = (Tree[root].left + Tree[root].right) >> 1;
string ret = "114514"; //************可能是 （-INF）要尽可能的小
if (end <= mid)
ret = max(ret, RmaxQ(root << 1, start, end)); //完全左孩子区间匹配
else if (start >= mid + 1)
ret = max(ret, RmaxQ((root << 1) + 1, start, end)); //完全右孩子区间匹配
else
{
string a = RmaxQ(root << 1, start, mid);
string b = RmaxQ((root << 1) + 1, mid + 1, end);
ret = max(a, b); //求的左右两个区间和匹配区间相符的最大值得较大者
}
return ret; //记得返回结果
}
int main() {
ios::sync_with_stdio(false);
cin.tie(nullptr);
int n, q;
cin >> n >> q;
for (int i = 1; i <= n; i ++)
cin >> A[i];
Build(1, 1, 1e5 + 1);
while (q--) {
char op ;
int x, y;
string s;
cin >> op ;
if (op == 'Q') {
cin >> x >> y;
if (RmaxQ(1, x, y) == "114514") cout << "null\n";
else cout << RmaxQ(1, x, y) << '\n';
} else {
cin >> x >> s;
update(1, x, s);
}
}
return 0;
}
