QBXT模拟赛1

QBXT模拟赛1

总结

期望得分：\(100 + 80 + 10 = 190\)
实际得分：\(90 + 80 + 10 = 180\)

这是在清北的第一场考试，也是在清北考的最高的一次了吧。。本来以为能拿\(190\)的，没想到强者太多，\(AK\)的一群，\(200\)分大众分。。我好菜

思路&&代码

T1

\(T1\)是个简单题，却因为\(1-1=0\)这个点忘记去除前导零而失去了\(10\)分，以后要多对拍，多注意细节

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
using namespace std;

const int N = 1e5 + 11;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

char s[N];
string a;
int len, now, whe, pos;

int main() {
	scanf("%s", s + 1);
	len = strlen(s + 1), now = s[1] - '0', whe = 1, pos = 1; 
	for(int i = 2, x; i <= len; i++) {
		x = s[i] - '0';
		if(x > now) {
			now = x;
			whe = i;
			pos = i;
		}
		else if(x == now) pos = i;
		if(x < now) break;
	}
	if(whe == len || pos == len) return cout << (s + 1) << '\n', 0;
	for(int i = 1; i < whe; i++) a += s[i];
	if(s[whe] - 1 != '0') a += s[whe] - 1;
	for(int i = whe + 1; i <= len; i++) a += '9';
	cout << a;
	return 0;
}

T2

看式子不懂，之后手算一下发现就是个逆序对，进而发现可以转化为求哪些区间包含这对逆序对，然后这对逆序对的值乘以区间个数，式子如下

\[\sum_j a_j * (n - j + 1) *\sum_{a_i > a_j, i < j} a_i * i \]

后面的可以用数据结构维护，发现模数是\(1e12+7\),两个\(10^12\)的数相乘会爆\(long\ long\)，所以要用快速乘

然后就做完了

#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define lowbit(x) (x & -x)
#define int long long
using namespace std;

const int N = 5e5 + 11;
const int mod = 1e12 + 7;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

int n;
int a[N], ans = 0;
int b[N], p[N], f[N];

inline int mul(int a, int b, int res = 0) {
	while(b) {
		if(b & 1) ans = (ans + a) % mod;
		a = a + a % mod; b >>= 1;
	}
	return res;
}

inline int query(int x) {
	int ans = 0;
	for(int i = x; i; i -= lowbit(i)) ans = (ans + p[i]) % mod;
	return ans;
}

const int MAX = 1e5;

inline void add(int x, int val) {
	for(int i = x; i <= MAX; i += lowbit(i)) p[i] = (p[i] + val) % mod;
	return;
}

signed main() {
	n = read();
	for(int i = 1; i <= n; i++) b[i] = a[i] = read();
	sort(b + 1, b + 1 + n);
	for(int i = 1; i <= n; i++) f[i] = lower_bound(b + 1, b + 1 + n, a[i]) - b;
	for(int i = 1; i <= n; i++) {
		int now = query(n) - query(f[i]);
		now = (now % mod + mod) % mod;
		ans = ans + mul(now * (n - i + 1), a[i]);
		ans = (ans % mod + mod) % mod;
		add(f[i], a[i] * i);
	}
	ans = (ans % mod + mod) % mod;
	cout << ans << '\n';
	return 0;
}

T3

直接用线段树扫描线就\(over\)了

还有一种神奇做法。。

#include <map>
#include <cmath>
#include <cstdio>
#include <cstring>
#include <iostream>
#include <algorithm>
#define max(a, b) (a > b ? a : b)
#define min(a, b) (a < b ? a : b)
#define PII pair<int, int>
#define mk(x, y) make_pair(x, y)
using namespace std;

const int N = 5e4 + 11;
const int M = 1e6 + 11;
const int mod = 1e9 + 7;
const int INF = 0x3f3f3f3f;

inline int read() {
	char c = getchar();
	int x = 0, f = 1;
	for( ; !isdigit(c); c = getchar()) if(c == '-') f = -1;
	for( ; isdigit(c); c = getchar()) x = (x << 3) + (x << 1) + (c ^ 48);
	return x * f;
}

int n, m, a[N], b[N], c[N], d[N];
int mina, minb, maxc, maxd;

map<pair<int, int>, int> mp;

int main() {
	int T = read();
	while(T--) {
		n = read();
		mina = minb = INF;
		maxc = maxd = -INF;
		mp.clear();
		for(int i = 1; i <= n; i++) {
			a[i] = read(), b[i] = read(), c[i] = read(), d[i] = read();
			mina = min(mina, a[i]);
			minb = min(minb, b[i]);
			maxc = max(maxc, c[i]);
			maxd = max(maxd, d[i]);
			mp[mk(a[i], b[i])]++;
			mp[mk(a[i], d[i])]++;
			mp[mk(c[i], b[i])]++;
			mp[mk(c[i], d[i])]++;
		}
		int cnt = 0;
		if(mp[mk(mina, minb)] == 1) cnt++;
		if(mp[mk(mina, maxd)] == 1) cnt++;
		if(mp[mk(maxc, minb)] == 1) cnt++;
		if(mp[mk(maxc, maxd)] == 1) cnt++;
		if(cnt != 4) {
			puts("Guguwansui");
			continue;
		}
		cnt = 0;
		for(int i = 1; i <= n; i++) {
			if(mp[mk(a[i], b[i])] == 1) cnt++;
			if(mp[mk(a[i], d[i])] == 1) cnt++;
			if(mp[mk(c[i], b[i])] == 1) cnt++;
			if(mp[mk(c[i], d[i])] == 1) cnt++;
		}
		if(cnt == 4) puts("Perfect");
		else puts("Guguwansui");
	}
	return 0;
}
//这题太神了我不会。