Codeforces Round #618 (Div. 2)

题目链接：https://codeforces.com/contest/1300

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A：

贪心

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 101;
int n, a[maxn];

int main() {
    int t; scanf("%d", &t);
    while (t--) {
        int ans = 0, sum = 0;
        scanf("%d", &n);
        for (int i = 1; i <= n; i++) {
            scanf("%d", &a[i]);
            if (!a[i]) {
                a[i] = 1, ans++;
            }
            sum += a[i];
        }
        if (!sum) ans++;
        printf("%d\n", ans);
    }
    return 0;
}

B：

白给

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 2e5 + 10;
int n, a[maxn];

int main() {
    int t; scanf("%d", &t);
    while (t--) {
        scanf("%d", &n);
        n = n * 2;
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        sort(a + 1, a + 1 + n);
        printf("%d\n", a[n / 2 + 1] - a[n / 2]);
    }
    return 0;
}

C：

从高到低地检查每一个数的每一位，找出满足如下条件的一个数：该数的某一位只在该数出现过一次，其他数没有。如果存在这样的数，就把它放到数组最前面并输出整个数组；否则直接输出整个数组。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define dou double
#define pb emplace_back
#define mp make_pair
#define sot(a,b) sort(a+1,a+1+b)
#define rep1(i,a,b) for(int i=a;i<=b;++i)
#define rep0(i,a,b) for(int i=a;i<b;++i)
#define eps 1e-8
#define int_inf 0x3f3f3f3f
#define ll_inf 0x7f7f7f7f7f7f7f7f
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 1e5 + 10;
int n, a[maxn];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
    for (int j = 30; ~j; j--) {
        int cnt = 0, pos;
        for (int i = 1; i <= n; i++)
            if ((a[i] >> j) & 1) {
                cnt++;
                pos = i;
            }
        if (cnt == 1) {
            printf("%d ", a[pos]);
            for (int i = 1; i <= n; i++) {
                if (i != pos) printf("%d ", a[i]);
            }
            puts("");
            return 0;
        }
    }
    for (int i = 1; i <= n; i++) printf("%d ", a[i]);
    puts("");
    return 0;
}

D：

题目有点长但其实很简单的一道题。给定一个凸包S，定义凸包T为凸包S能覆盖原点的区域(可以想象到T就是S在原点周围平移的可行域)，问S与T是否相似。

满足相似的充要条件是凸包S点数为偶数，且S为中心对称图形。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define pb emplace_back
#define mp make_pair
#define eps 1e-8
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 1e5 + 10;
map<pair<ll, ll>, int>m;
pair<ll, ll>a[maxn];
int n;
ll midX = 0, midY = 0;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        scanf("%lld%lld", &a[i].first, &a[i].second);
        // 放缩坐标，防止出现小数
        a[i].first *= n, a[i].second *= n;
        midX += a[i].first, midY += a[i].second;
        m[mp(a[i].first, a[i].second)] = 1;
    }
    midX /= n, midY /= n;
    for (int i = 1; i <= n; i++) {
        if (!m.count(mp(2 * midX - a[i].first, 2 * midY - a[i].second))) {
            return puts("NO"), 0;
        }
    }
    puts("YES");
    return 0;
}

E：

做法比较贪心。例如现在手上有一个已经处理好的数组[x1,x1,...,x1][x2,x2,...,x2]，第一块平均值为x1，第二块平均值为x2,。倘若在该数组后面加上一个新数x3，那么就需要比较x3跟x2的大小关系。如果x3<x2，就肯定要把x3合并到前一块里，再比较第二块新平均值与x1的大小关系，再决定是否要进行比较。

写起来像O(n^2)，但其实每合并一块就会使得块的数量-1，所以总体复杂度为均摊O(n)。

/* basic header */
#include <bits/stdc++.h>
/* define */
#define ll long long
#define pb emplace_back
#define mp make_pair
#define eps 1e-8
#define lson (curpos<<1)
#define rson (curpos<<1|1)
/* namespace */
using namespace std;
/* header end */

const int maxn = 1e6 + 10;
int n, p;
double a[maxn], len[maxn], ans[maxn];

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) scanf("%lf", &a[i]);
    for (int i = 1; i <= n; i++) {
        ans[++p] = a[i];
        len[p] = 1;
        while (p > 1 && ans[p - 1] > ans[p]) {
            ans[p - 1] = (ans[p] * len[p] + ans[p - 1] * len[p - 1]) / (len[p] + len[p - 1]);
            len[p - 1] += len[p];
            p--;
        }
    }
    for (int i = 1; i <= p; i++) {
        for (int j = 1; j <= len[i]; j++) {
            printf("%.12f ", ans[i]);
        }
    }
    return 0;
}