Codeforces Round #Pi (Div. 2)

题目传送：Codeforces Round #Pi (Div. 2)

A. Lineland Mail

水题。。

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n;

int a[100005];

int main() {
    scanf("%d", &n);
    for(int i = 0; i < n; i ++) {
        scanf("%d", &a[i]);
    }
    for(int i = 0; i < n; i ++) {
        int ma = a[i] - a[0], mi = INF;
        ma = max(ma, a[n-1] - a[i]);
        if(i - 1 >= 0 && i - 1 < n) {
            mi = min(mi, a[i] - a[i-1]);
        }
        if(i + 1 >= 0 && i + 1 < n) {
            mi = min(mi, a[i + 1] - a[i]);
        }
        printf("%d %d\n", mi, ma);
    }
    return 0;
}

---

B. Berland National Library

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

set<int> s;

map<int, int> mp;

char str[10];
int a;

int main() {
    int n;
    int jin_max = 0;//维护一个通过'+'而进阅读室的最大值
    int chu_max = 0;//维护一个之前就在阅读室的最大值，则jin_max+chu_max为答案
    scanf("%d", &n);

    for(int i = 0; i < n; i ++) {
        scanf("%s %d", str, &a);
        if(str[0] == '+') {//进
            s.insert(a);//放到一个集合里面
            mp[a] = 1;//标记为访问过
            jin_max = max(jin_max + chu_max, (int)s.size()) - chu_max;//维护一下进去的最大值
        }
        else {//出
            if(mp.find(a) == mp.end()) {//如果之前都没出现过，那说明刚开始就在阅读室。
                chu_max ++;
            }
            else {
                s.erase(a);//之前出现过，那么现在现在肯定在set里面，出set
            }
        }
//      cout << jin_max + chu_max << endl;
    }

    cout << jin_max + chu_max << endl;
    return 0;
}

---

C. Geometric Progression

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

const int maxn = 200005;
LL zuo[maxn];//维护一个值i左边的i/k的个数
LL you[maxn];//维护一个值i右边的i*k的个数（注意这里可能会爆int，又跪了TOT）
LL a[maxn];

mapint> mp;//对应从左往右遍历
mapint> mp2;//对应从右往左遍历

int n, k;

int main() {
    scanf("%d %d", &n, &k);
    for(int i = 0; i < n; i ++) {
        scanf("%I64d", &a[i]);

        if(a[i] % k == 0) {
            if(mp.find(a[i] / k) != mp.end()) {
                zuo[i] = mp[a[i] / k];
            }
            else {
                zuo[i] = 0;
            }
        }
        else {
            zuo[i] = 0;
        }

        if(mp.find(a[i]) == mp.end()) mp[a[i]] = 1;
        else {
            mp[a[i]] ++;
        }


    }

    for(int i = n - 1; i >= 0; i --) {

        if(mp2.find(a[i] * k) != mp2.end()) {
            you[i] = mp2[a[i] * k];
        }
        else {
            you[i] = 0;
        }

        if(mp2.find(a[i]) == mp2.end()) mp2[a[i]] = 1;
        else {
            mp2[a[i]] ++;
        }

    }

    LL ans = 0;
    for(int i = 0; i < n; i ++) {
        ans += zuo[i] * you[i];
//      cout << zuo[i] << " " << you[i] << endl;
    }

    cout << ans << endl;

    return 0;
}

---

D. One-Dimensional Battle Ships

set里可以只记录shot的位置，再另加0和n+1这两个边界

而之前存的每个区间太慢了。。完美超时。。

AC代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, k, a;
int m;

int fun(int x) {//算出一个区间可以放几条船
    return (x + 1) / (a + 1);
}

int main() {
    map<int, int> mp;
    scanf("%d %d %d", &n, &k, &a);
    scanf("%d", &m);
    mp[0] = mp[n + 1] = 1;

    int ship_sum = fun(n);

    for(int i = 1; i <= m; i ++) {
        int shot;
        scanf("%d", &shot);

        if(mp.find(shot) != mp.end()) {
            continue;
        }
        mp[shot] = 1;

        map<int, int>::iterator low = mp.lower_bound(shot);
        low --;
        map<int, int>::iterator up = mp.upper_bound(shot); 

        int l = low->first;
        int r = up->first;

        ship_sum -= fun(r - l - 1);
        ship_sum += fun(shot - l - 1) + fun(r - shot - 1);

        if(ship_sum < k) {
            cout << i << endl;
            return 0;
        }

    }
    cout << -1 << endl;
    return 0;
}

TLE的代码：

#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#include 
#define LL long long
#define INF 0x7fffffff
using namespace std;

int n, k, a;
int m, t;

map<int, int> mp;

struct node {
    int l, r;
    node() {
    }
    node(int _l, int _r) : l(_l), r(_r) {
    }
    bool operator < (const node& a) const {
        return l < a.l;
    }
};

//bool operator < (node &a, node &b) {
//  return a.l < b.l;
//}

set s;

bool fun() {
    int cnt = 0;
    for(set::iterator it = s.begin(); it != s.end(); it ++) {
//      cout << it->r - it->l + 1 << " ";
        cnt += ( (it->r - it->l + 2) / (a + 1) );
    }
//  cout << endl;
    if(cnt >= k) return true;
    else return false;
}

int main() {
    scanf("%d %d %d", &n, &k, &a);
    s.insert(node(1, n));
    scanf("%d", &m);
    int ans;
    int flag = 0;
    for(int i = 0; i < m; i ++) {
        scanf("%d", &t);

        if(mp.find(t) == mp.end()) {
            mp[t] = 1;
        }
        else continue;

        if(flag == 1) continue;

        set::iterator it = s.upper_bound(node(t, t));
        it --;
        int rr = it->r;
        int ll = it->l;
        s.erase(it);
        if(ll == rr && rr == t) {
        }
        else if(rr == t) {
            s.insert(node(ll, rr - 1));
        }
        else if(ll == t) {
            s.insert(node(t + 1, rr));
        }
        else {
            s.insert(node(ll, t - 1));
            s.insert(node(t + 1, rr));
        }
        if(!fun()) {
            flag = 1;
            ans = i + 1;
        }
    }
    if(flag == 0) {
        printf("-1\n");
    }
    else printf("%d\n", ans);
    return 0;
}