Codeforces Round #Pi (Div. 2)

A - Lineland Mail

水题，拼手速。

/*
* @author Novicer
* language : C++/C
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
#define INF 2147483647
#define cls(x) memset(x,0,sizeof(x))
#define rise(i,a,b) for(int i = a ; i <= b ; i++)
using namespace std;
const double eps(1e-8);
typedef long long lint;

const int maxn = 100000 + 5;
lint x[maxn];
int main(){
	int n;
	cin >> n;
	for(int i = 1 ; i <= n ; i++)	
		scanf("%I64d",&x[i]);
	sort(x+1,x+n+1);//事实上不用排序的:)
	cout << x[2] - x[1] << " " << x[n] - x[1] << endl;
	for(int i = 2 ; i < n ; i++){
		lint mi = min(x[i] - x[i-1] , x[i+1] - x[i]);
		lint ma = max(x[n] - x[i] , x[i] - x[1]);
		printf("%I64d %I64d\n",mi,ma);
	}
	cout << x[n] - x[n-1] << " " << x[n] - x[1] << endl;
	return 0;
}

B - Berland National Library

预处理一開始没出现过的“- x”，在开头补充“+ x"

然后用set模拟进入与离开的过程。输出最大的set容量。

/*
* @author Novicer
* language : C++/C
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
#define INF 2147483647
#define cls(x) memset(x,0,sizeof(x))
#define rise(i,a,b) for(int i = a ; i <= b ; i++)
using namespace std;
const double eps(1e-8);
typedef long long lint;
bool vis[1000005];
int del[105],add[105],s[105];
set<int>lib;
int main(){
//	freopen("input.txt","r",stdin);
	int n;
	cin >> n;
	cls(vis);
	cls(del);cls(add);cls(s);
	int ans = 0;
	int tmp;
	int x = 0;
	int ans1 = 0 , ans2 = 0;
	for(int i = 1 ; i <= n ; i++){
		char c;
		cin >> c;
		if(c == ‘+‘){
			cin >> add[i];
			vis[add[i]] = true;
		}
		else{
			int num;
			cin >> num;
			if(!vis[num]){
				x++;
				s[x] = num;
				del[i] = num;
			}
			else{
				del[i] = num;
			}
		}
	}
	for(int i = 1 ; i <= x ; i++) lib.insert(s[i]);
	ans = lib.size();
	for(int i = 1 ; i <= n ; i++){
		if(add[i] != 0){
			lib.insert(add[i]);
		}
		else if(del[i] != 0){
			lib.erase(del[i]);
		}
		tmp = lib.size();
		ans = max(tmp , ans);
	}
	cout << ans << endl;
	return 0;
}

C - Geometric Progression 

枚举每个元素a[i]。将其当做等比数列的第二项，假设a[i]/k 与 a[i]*k存在。则将a[i]/k 与 a[i]*k的数量相乘并累加。

英文题解好像解释的更清楚些。。

Let‘s solve this problem for fixed middle element of progression. This means that if we fix element a_i then the progression must consist of a_i?/?k and a_i·k elements. It could not be possible, for example, if a_i is not divisible by k ().

For fixed middle element one could find the number of sequences by counting how many a_i?/?k elements are placed left from fixed element and how many a_i·k are placed right from it, and then multiplying this numbers. To do this, one could use two associative arrays A_l and A_r, where for each key x will be stored count of occurences of x placed left (or right respectively) from current element. This could be done with map structure.

Sum of values calculated as described above will give the answer to the problem.

/*
* @author Novicer
* language : C++/C
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
#define INF 2147483647
#define cls(x) memset(x,0,sizeof(x))
#define rise(i,a,b) for(int i = a ; i <= b ; i++)
using namespace std;
const double eps(1e-8);
typedef long long lint;

const int maxn = 2 * 100000 + 5;
lint a[maxn];
map<lint , lint> l;
map<lint , lint> r;
int main(){
//	freopen("input.txt","r",stdin);
	int n;
	lint k;
	cin >> n >> k;
	for(int i = 1 ; i<= n ; i++){
		scanf("%I64d",&a[i]);
		r[a[i]]++;
	}
	lint ans = 0;
	for(int i = 1 ; i <= n ; i++){
		r[a[i]]--;
		if(a[i] % k == 0)
			ans += l[a[i]/k] * r[a[i]*k];
		l[a[i]]++;
	}
	cout << ans << endl;
	return 0;
}

D - One-Dimensional Battle Ships 

非常有意思的一道题，长度为n的空间放着k条长度为a的船，且船不重叠，Alice给你m个宣称miss（即没有船）的位置。推断Alice是否撒谎，若撒谎输出第一次能够发现矛盾的位置。

思路：

题解思路太棒！

简单来说就是推断Alice说的位置将空间分为两个部分，计算出这两个部分各能容纳几艘船，加起来与k比較，大于则撒谎。

/*
* @author Novicer
* language : C++/C
*/
#include<iostream>
#include<sstream>
#include<fstream>
#include<vector>
#include<list>
#include<deque>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<bitset>
#include<algorithm>
#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cctype>
#include<cmath>
#include<ctime>
#include<iomanip>
#define INF 2147483647
#define cls(x) memset(x,0,sizeof(x))
#define rise(i,a,b) for(int i = a ; i <= b ; i++)
using namespace std;
const double eps(1e-8);
typedef long long lint;

map<int , int> ship;
int n,k,a;

int f(int x){return (x+1)/(a+1);};//计算长度为x区间能容纳几艘船
int main(){
//	freopen("input.txt","r",stdin);
	while(cin >> n >> k >> a){
		ship.clear();
		int m ; cin >> m;
		ship[0] = ship[n+1] = 1;
		int max_sum = f(n);
		bool cheat = false;
		for(int i = 1 ; i <= m ; i++){
			int shot;
			scanf("%d",&shot);
			if(ship.find(shot) != ship.end()) continue;
			ship[shot] = 1;
			int l , r;
			map<int , int>:: iterator low = ship.lower_bound(shot);
			map<int , int>:: iterator up = ship.upper_bound(shot);//low-1与up各自是最靠近shot的左右位置
			l = (--low)->first;
			r = up->first;
			max_sum -= f(r - l - 1);
			max_sum += f(r - shot - 1) + f(shot - l - 1);
			if(max_sum < k){
				cheat = true;
				cout << i << endl;
				return 0;
			}
		}
		if(!cheat)
			cout << -1 << endl;
	}
	return 0;
}