第三届中国计量大学ACM程序设计竞赛个人赛

链接
A:模拟

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e5+10;
const int mod = 1e9+7;

string s;
stack sta;
int main(){
	getline(cin, s);
	string str = "";
	for(int i = 0; i < s.size(); i++){
		if(s[i] == ' ') continue;
		if(s[i] == '-') s[i] = '*';
		else if(s[i] == '/') s[i] = '+';
		else if(s[i] == '+') s[i] = '/';
		else if(s[i] == '*') s[i] = '-';
		str += s[i];
	}
	//cout << str << "\n";
	int p = 0, op = 1, flag = 0;
	for(int i = 0; i <= str.size(); i++){
		if(str[i] <= '9' && str[i] >= '0' && i < str.size()){
			p = p*10 + op * (str[i] - '0');
		}
		else{
			if(flag == 1){
				double q = sta.top();
				sta.pop();
				sta.push(p*q);
				//cout << p << " " << q << "**\n";
			}
			else if(flag == 2){
				if(p == 0){
					printf("Cannot be divided by 0\n");
					return 0;
				}
				double q = sta.top();
				sta.pop();
				sta.push(q/p);
				//cout << q << " " << p << "\n";
			}
			else {
				sta.push(p*1.0);
				//cout << p << "\n";
			}
			if(i == str.size()) break;
			p = 0; op = 1; flag = 0;
			if(str[i] == '-') op = -1;
			else if(str[i] == '*') flag = 1;
			else if(str[i] == '/') flag = 2;
		}
	}
	double ans = 0;
	while(!sta.empty()){
		double q = sta.top();
		sta.pop();
		ans += q;
	}
	printf("%.2lf\n", ans);
	return 0;
}

B:

假设$min(x_a+y_b,x_b+y_a)=x_a+y_b$, $min(x_a+y_c,x_c+y_a)=x_a+y_c$, 要使$x_a+y_b>x_a+y_c$, 则$y_b>y_c$
综上所述，当$x_a+y_b>x_b+y_a,x_a+y+c>x_b+y_a，y_b>y_c$时，选a,b更优，所以按照x_a-y_a从大到小排序，对每个x选前面最大的y形成最优解，取max

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 2e5+10;
const int mod = 1e9+7;

int n;
struct node{
	int a, b, pos;
	bool operator < (const node& e) const {
		return a-b > e.a - e.b;
	}
}e[N];


int main(){
	IO;
	cin >> n;
	for(int i = 0; i < n; i++){
		cin >> e[i].a;
		e[i].pos = i;
	}
	for(int i = 0; i < n; i++){
		cin >> e[i].b;
	}
	sort(e, e+n);
	node no = e[0];
	int ans = 0;
	for(int i = 1; i < n; i++){
		ans = max(ans, e[i].a + no.b);
		if(e[i].b > no.b){
			no = e[i];
		}
	}
	cout << ans <<"\n";
	return 0;
}

D:水题

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e5+10;
const int mod = 1e9+7;

int t, x, h, m;

int main(){
	cin >> t;
	while(t--){
		scanf("%d", &x);
		scanf("%d:%d", &h, &m);
		int tm = h*60 + m;
		int y;
		for(int i = 0; i <= 1e6; i++){
			if(tm%10 == 7 || tm%60/10 == 7 || tm/60%10 == 7){
				y = i;
				break;
			}
			tm -= x;
			if(tm < 0){
				tm += 24*60;
			}
		}
		tm %= (24*60);
		h = tm/60, m = tm%60;
		printf("%d %02d:%02d\n", y, h, m);
	}
	return 0;
}

L:

思路来自
其实感觉这题比较简单，但是过题人数好像不太多，我们先思考一下不带修改的括号匹配的过程，很明显我们可以定义一个cnt遇到 ( 就加一，遇到 ) 就减一，然后只要整个过程中cnt都大于等于零，并且最后cnt等于0就说明括号匹配是成功的。那么我们想按照题目修改后会怎么样，其实很容易发现，如果将当前位置的 ( 变为 ) 实际上就是将之后的cnt全部减二，同理如果 ) 变为 ( 其实就是将之后的cnt全部加二。那么我们很容易发现上面的过程和判断包含了区间修改，区间查询和单点查询，所以我们只要用线段树维护修改的过程，每次查询整体的最小是不是大于等于0，并且最后等于0就行了。

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e5+10;
const int mod = 1e9+7;

int n, m;
char s[N];
int cnt[N], Min[N*4], lazy[N*4];

void build(int o, int l, int r){
	if(l == r){
		Min[o] = cnt[l];
		return ;
	}
	int mid = l+r>>1;
	build(o<<1, l, mid);
	build(o<<1|1, mid+1, r);
	Min[o] = min(Min[o<<1], Min[o<<1|1]);
}

void push(int o){
	if(lazy[o]){
		lazy[o<<1] += lazy[o];
		lazy[o<<1|1] += lazy[o];
		Min[o<<1] += lazy[o];
		Min[o<<1|1] += lazy[o];
		lazy[o] = 0;
	}
}

void up(int o, int l, int r, int ql, int qr, int v){
	if(l >= ql && r <= qr){
		Min[o] += v;
		lazy[o] += v;
		return ;
	}
	push(o);
	int mid = l+r>>1;
	if(mid >= ql) up(o<<1, l, mid, ql, qr, v);
	if(mid < qr) up(o<<1|1, mid+1, r, ql, qr, v);
	Min[o] = min(Min[o<<1], Min[o<<1|1]);
}

int qu(int o, int l, int r, int ql, int qr){
	if(l >= ql && r <= qr){
		return Min[o];
	}
	push(o);
	int res = 1e9, mid = l+r>>1;
	if(mid >= ql) res = min(res, qu(o<<1, l, mid, ql, qr));
	if(mid < qr) res = min(res, qu(o<<1|1, mid+1, r, ql, qr));
	Min[o] = min(Min[o<<1], Min[o<<1|1]);
	return res;
}

int main(){
	IO;
	cin >> n >> m;
	cin >> s+1;
	cnt[0] = 0;
	for(int i = 1; i <= n; i++){
		if(s[i] == '('){
			cnt[i] = cnt[i-1] + 1;
		}
		else cnt[i] = cnt[i-1] - 1;
	}
	build(1, 1, n);
	int x;
	while(m--){
		cin >> x;
		if(s[x] == '('){
			s[x] = ')';
			up(1, 1, n, x, n, -2);
		}
		else{
			s[x] = '(';
			up(1, 1, n, x, n, 2);
		}
		int ans1 = qu(1, 1, n, 1, n);
		int ans2 = qu(1, 1, n, n, n);
		if(ans1 >= 0 && ans2 == 0){
			cout << "Yes\n";
		}
		else cout << "No\n";
	}
	return 0;
}

N:

#include
using namespace std;
typedef long long ll;
#define IO ios::sync_with_stdio(false)
#define pb push_back
#define mk make_pair
const int N = 1e6+10;
const int mod = 1e9+7;
int t, n;
ll f[N];

int main(){
	IO;
	f[1] = 2;
	for(int i = 2; i <= 1e6; i++){
		f[i] = (f[i-1]*3%mod + 2)%mod;
	}
	cin >> t;
	while(t--){
		cin >> n;
		cout << f[n] << "\n";
	}
	return 0;
}