Codeforces Round #Pi (Div. 2) (STL專場)
Codeforces Round #Pi (Div. 2)
水題，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);//事實上不用排序的:)
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
預處理一開始沒出現過的“- x”，在開頭補充“+ x"
然後用set模擬進入與離開的過程。輸出最大的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
枚舉每個元素a[i]。將其當做等比數列的第二項，假設a[i]/k 與 a[i]*k存在。則將a[i]/k 與 a[i]*k的數量相乘並累加。
英文題解好像解釋的更清楚些。。
Let‘s solve this problem for fixed middle element of progression. This means that if we fix element
ai then the progression must consist of
ai?/?k and
ai·k elements. It could not be possible, for example, if
a
).
For fixed middle element one could find the number of sequences by counting how many ai?/?k elements are placed left from fixed element and how many ai·k are placed right from it, and then multiplying this numbers. To do this, one could use two associative arrays Al and Ar, 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
非常有意思的一道題，長度為n的空間放著k條長度為a的船，且船不重疊，Alice給你m個宣稱miss（即沒有船）的位置。推斷Alice是否撒謊，若撒謊輸出第一次能夠發現矛盾的位置。
思路：
題解思路太棒！
簡單來說就是推斷Alice說的位置將空間分為兩個部分，計算出這兩個部分各能容納幾艘船，加起來與k比較，大於則撒謊。
For problem D, simply use a map structure to record the free space (a consecutive empty square without shotpoint) on a shotpoint‘s left and right.
You can use the lower_bound and upper_bound functions to find the nearest shotpoint, and update them.
When a shot happens a free space will be broken into 2, so calculate the sum of max number of ships of those 2 spaces and subtract it from the global sum of max number of ships.
Calculate the maximum number of ships could be placed in the space: (space+1)/(a+1)
(global sum of max number of ships = (n+1)/(a+1) before the 1st shot.)
when the global sum < k just print the current operation number and exit.
/*
* @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);};//計算長度為x區間能容納幾艘船
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與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;
}
Codeforces Round #Pi (Div. 2) (STL專場)