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2025-10-25 MXOJ 【CSP-S】-- 模拟四【郑州一中】record

news 2025/10/28 13:34:29

比赛链接：Contests

A.put

原题链接：Problems

简单的贪心，直接切了。10点58分榜上的rank1大佬写了个dp，但我研究了一下，那个dp也只是把贪心的过程记录在了dp数组中。

那么直接上代码吧！

正解：

#include <bits/stdc++.h>
using namespace std;
const int N = 200005;
long long n;
struct node{long long p, h;
}a[N];
bool cmp(node x, node y){return x.p < y.p;
}
int read(){char c = getchar();long long x = 0, f = 1;while (c < '0' || c > '9'){if (c == '-')f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = x * 10 + (c - '0');c = getchar();}return x * f;
}
int main(){n = read();for (int i = 1; i <= n; i++){a[i].p = read();a[i].h = read();}sort(a + 1, a + n + 1, cmp);long long ans = 2;for (int i = 2; i < n; i++){int lst = a[i - 1].p;int nxt = a[i + 1].p;if (a[i].p - a[i].h > lst){ans++;continue;}if (a[i].p + a[i].h < nxt){ans++;a[i].p += a[i].h;}}printf("%d", ans);
}

常数很大吗？机房普遍都在100ms以内，我跑了131ms/(ㄒoㄒ)/~~

B.match

原题链接：Problems

错误分析：乱打暴力，写了 $O(n^{2}log^{2}n)$ ,其实思考一下就可以优化到 $O\left ( nlogn \right )$

思路分析：相当于把A数组进行滑动，而前面的是已经存在的，只改变一个位置就可以了。

正解：

#include<bits/stdc++.h>
#define ll long long
using namespace std;
const ll N = 2e5 + 10;
ll n, m, q, p;
ll a[N], o[N], mx[N], ans[N];
struct node{ll num, w;
}b[N];
bool cmp(node q, node p){return q.num < p.num;
}
long long read(){char c = getchar();long long x = 0, f = 1;while (c < '0' || c > '9'){if (c == '-')f = -1;c = getchar();}while (c >= '0' && c <= '9'){x = x * 10 + (c - '0');c = getchar();}return x * f;
}
int main(){    n = read();n++;for (ll i = 1; i <= n; i++){b[i].num = read();b[i].w = i;}for (ll i = 1; i < n; i++){a[i] = read();}sort(a + 1, a + n);sort(b + 1, b + 1 + n, cmp);for (ll i = n; i >= 2; i--){o[i] = max(o[i + 1], b[i].num - a[i - 1]);}ans[b[1].w] = o[2];for (ll i = 2; i <= n; i++){mx[i] = max(b[i - 1].num - a[i - 1], mx[i - 1]);ans[b[i].w] = max(mx[i], o[i + 1]);}for (ll i = 1; i <= n; i++){printf("%lld ", ans[i]);}return 0;
}

膜拜呱呱！

C.div

显然我只会 $O(n)$ ，但要求 $O(\sqrt{n})$ .

传言是根号分治的板子，机房AeeE5x这么强吗？

那么，直接上正解吧！！！

#include<bits/stdc++.h>
#define int long long
using namespace std;
int prexor(int n){if(n == 2) return 3;if(n % 2 == 1) return ((n + 1) >> 1) % 2;else return n + ((n >> 1) % 2);
}
void solve(int n){int ans = 0;int nowfactor = 1;for ( ; nowfactor <= n; nowfactor++){if ((n / nowfactor) % 2 == 1) ans ^= nowfactor;elseans ^= 0;if (nowfactor > sqrt(n) && (n / nowfactor != n / (nowfactor + 1))) break; }nowfactor++;int maxdiv = n / nowfactor;for (int nowdiv = 1; nowdiv <= maxdiv; nowdiv += 2){int minfac = (n / (nowdiv + 1)) + 1;int maxfac = (n / nowdiv);ans ^= prexor(maxfac) ^ prexor(minfac - 1);}cout << ans << "\n";
}
int n;
signed main(){ios::sync_with_stdio(false);cin.tie(0),cout.tie(0);freopen("div.in", "r", stdin);freopen("div.out", "w", stdout);cin >> n;solve(n);return 0;
}

严肃学习根号分治……

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