Public NOIP Round1，2,3部分题解
1. Public NOIP Round #1 (Div. 1, 提高 2022-09-10 14:00:00)
A. 斜二等轴测图
就是个大模拟:将整个图形分为三个部分即可,简单计算亿下就可以了
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#include <stdio.h>
#include <string.h>
char map[1000][1000];
int main() {
int T, a, b, c;
scanf("%d", &T);
while(T --) {
scanf("%d%d%d", &a, &b, &c);
for(int i = 0; i < ((b + c) << 1 | 1); i ++)
for(int j = 0; j < ((a + b) << 1 | 1); j ++) map[i][j] = '.';
for(int i = 0; i < (b << 1); i ++) {
if(i & 1) {
for(int j = 0; j < (a << 1 | 1); j += 2) map[i][(b << 1) - i + j] = '/';
for(int j = 1; j < (a << 1 | 1); j += 2) map[i][(b << 1) - i + j] = '.';
for(int j = 0; j < (c << 1); j ++)
map[i + 1 + j][((a + b) << 1) - i] = j & 1 ? '/' : '.';
} else {
for(int j = 0; j < (a << 1 | 1); j += 2) map[i][(b << 1) - i + j] = '+';
for(int j = 1; j < (a << 1 | 1); j += 2) map[i][(b << 1) - i + j] = '-';
for(int j = 0; j < (c << 1); j ++)
map[i + 1 + j][((a + b) << 1) - i] = j & 1 ? '+' : '|';
}
}
for(int i = 0; i < (c << 1); i ++)
if(i & 1) {
for(int j = 0; j < (a << 1 | 1); j += 2) map[i + (b << 1)][j] = '|';
for(int j = 1; j < (a << 1 | 1); j += 2) map[i + (b << 1)][j] = '.';
} else {
for(int j = 0; j < (a << 1 | 1); j += 2) map[i + (b << 1)][j] = '+';
for(int j = 1; j < (a << 1 | 1); j += 2) map[i + (b << 1)][j] = '-';
}
for(int j = 0; j < (a << 1 | 1); j += 2) map[(b + c) << 1][j] = '+';
for(int j = 1; j < (a << 1 | 1); j += 2) map[(b + c) << 1][j] = '-';
for(int i = 0; i < ((b + c) << 1 | 1); i ++, putchar(10))
for(int j = 0; j < ((a + b) << 1 | 1); j ++) putchar(map[i][j]);
}
return 0;
}
先选所有横坐标相同的最左边或最右边
从(纵坐标)外往内枚举,如果个数>2则将纵坐标在中间的往横坐标移动,使用set维护即可
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#include <set>
#include <vector>
#include <stdio.h>
#include <string.h>
#include <algorithm>
const int N = 1e6 + 5;
typedef std::pair<int, int> PII;
int n;
struct P { int x, y, id; } a[N];
std::set<PII> set[N];
bool used[N];
std::vector<PII> rem;
int main() {
scanf("%d", &n);
for(int i = 0; i < n; i ++) scanf("%d%d", &a[i].x, &a[i].y), a[i].id = i;
std::sort(a, a + n, [](const P &x, const P &y) { return x.x < y.x || (x.x == y.x && x.y < y.y); });
for(int i = 0; i < n; i ++)
if((!i || a[i].x != a[i - 1].x) || (a[i].x != a[i + 1].x))
set[a[i].y].insert({a[i].x, i});
for(int t = 0; t < 4; t ++) {
for(int i = 1; i <= 1000000; i ++)
if(set[i].size() > 2U) {
for(auto it = ++ set[i].begin(), ed = -- set[i].end(); it != ed; it ++) {
int id = it->second;
if(a[id + 1].x == a[id].x)
set[a[id + 1].y].insert({a[id + 1].x, id + 1}), rem.push_back(*it);
}
while(rem.size()) set[i].erase(rem.back()), rem.pop_back();
}
for(int i = 1000000; i; i --)
if(set[i].size() > 2U) {
for(auto it = ++ set[i].begin(), ed = -- set[i].end(); it != ed; it ++) {
int id = it->second;
if(id && a[id - 1].x == a[id].x)
set[a[id - 1].y].insert({a[id - 1].x, id - 1}), rem.push_back(*it);
}
while(rem.size()) set[i].erase(rem.back()), rem.pop_back();
}
}
for(int i = 1; i <= 1000000; i ++)
if(set[i].size()) used[a[set[i].begin()->second].id] = used[a[set[i].rbegin()->second].id] = true;
for(int i = 0; i < n; i ++) printf("%d", int(used[i]));
putchar(10);
return 0;
}
2. Public NOIP Round #2 (Div. 1, 提高 2022-10-04 08:30:00)
A 恰钱
先打个表(递推或者dfs等枚举),然后使用lower_bound找到大于等于l的最小满足条件的数，判断这个数与r的大小关系
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#include <stdio.h>
#include <string.h>
#include <algorithm>
const int N = 1e7 + 5;
int num[N], top[N], tot; // top为最高位的1的位置
int T, l, r, res;
int main() {
num[++ tot] = 2, top[tot] = 1;
for(int i = 2; i <= 30; i ++) for(int j = 1; j <= tot; j ++) // 扩展
if(i > top[j] && !(num[j] << 1 >> i & 1) && (num[j] << 1 | (1 << i)) <= 1000000000)
num[++ tot] = num[j] << 1 | (1 << i), top[tot] = i;
scanf("%d", &T);
while(T --) {
scanf("%d%d", &l, &r);
res = *std::lower_bound(num + 1, num + tot + 1, l);
printf("%d\n", !res || res > r ? -1 : res);
}
return 0;
}
B 排序
使用DP: 设 f[i] 表示处理好 [i,n] 且 a[i] 必选的子序列长度的最大值
使用单调栈求出比这个数小的第一个数的下标 nxt[i]
则 f[i]<-f[j]+1 其中 j∈[nxt[i],nxt[nxt[i]])
答案为 max{f[j]} 其中 j∈[1,nxt[1]]
方便枚举: 将这些按照 a 排序
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#include <stdio.h>
#include <string.h>
#include <algorithm>
const int N = 5e5 + 5;
int n, ord[N], nxt[N], a[N], tr[N << 2];
inline int ls(int u) { return u << 1; }
inline int rs(int u) { return u << 1 | 1; }
void modify(int u, int l, int r, int x, int y) {
if(l == r) tr[u] = std::max(tr[u], y);
else {
int mid = (l + r) >> 1;
if(x <= mid) modify(ls(u), l, mid, x, y);
else modify(rs(u), mid + 1, r, x, y);
tr[u] = std::max(tr[ls(u)], tr[rs(u)]);
}
}
int query(int u, int l, int r, int x, int y) {
if(x <= l && r <= y) return tr[u];
int mid = (l + r) >> 1, res = 0;
if(x <= mid) res = std::max(res, query(ls(u), l, mid, x, y));
if(y > mid) res = std::max(res, query(rs(u), mid + 1, r, x, y));
return res;
}
int stk[N], top, res;
int main() {
scanf("%d", &n);
for(int i = 1; i <= n; i ++) scanf("%d", a + i), ord[a[i]] = i;
stk[top = 0] = n + 1, nxt[n + 1] = n;
for(int i = n; i; i --) {
while(top && a[i] > a[stk[top]]) top --;
nxt[i] = stk[top], stk[++ top] = i;
}
for(int i = n; i; i --)
modify(1, 1, n, ord[i], query(1, 1, n, nxt[ord[i]], nxt[nxt[ord[i]]] - 1) + 1);
printf("%d\n", query(1, 1, n, 1, nxt[1]));
return 0;
}
1. Public NOIP Round #3 (Div. 1, 提高 2022-10-22 08:30:00) 题解
| A | 移除石子 |90/
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#include <stdio.h>
#include <string.h>
#include <algorithm>
#define x first
#define y second
const int N = 3005;
int n;
std::pair<int, int> a[N];
int main() {
int T;
scanf("%d", &T);
while(T --) {
scanf("%d", &n);
for(int i = 1; i <= n; i ++) scanf("%d%d", &a[i].first, &a[i].second);
std::sort(a + 1, a + n + 1), puts("Yes");
while(n) {
auto X = a[n], Y = a[n - 1];
if(X.x == Y.x) printf("%d %d %d %d\n", Y.x, Y.y, Y.x + X.y - Y.y, X.y);
else if(Y.y <= X.y) {
int t = std::max(X.x - Y.x, X.y - Y.y);
printf("%d %d %d %d\n", Y.x, Y.y, Y.x + t, Y.y + t);
} else {
auto p = [&] () {
int t = std::max(X.x - Y.x, Y.y - X.y);
printf("%d %d %d %d\n", Y.x, X.y, Y.x + t, X.y + t);
};
if(n > 2) {
auto Z = a[n - 2];
if(Z.x == Y.x) {
if(Z.y > X.y) printf("%d %d %d %d\n", Z.x, Z.y, Z.x + Y.y - Z.y, Y.y), a[n - 2] = X;
else if(Z.y == X.y) {
int dx = X.x - Z.x, dy = Y.y - Z.y;
if(dx >= dy) printf("%d.5 %d %d.5 %d\n", Z.x - 1, Z.y, Z.x + dy - 1, Y.y), a[n - 2] = X;
else if(dx < dy) printf("%d %d %d %d\n", Z.x, Z.y, X.x, Z.y + dx), a[n - 2] = Y;
} else p();
} else p();
} else p();
}
n -= 2;
}
}
return 0;
}
| B | 抓内鬼 |/20
| C | 异或序列 |/30
