Codeforces Round #579 (Div. 3)

打得巨差。题目链接：https://codeforces.com/contest/1203

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A：

正着来一遍倒着来一遍就行。

/* 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 */

int q;
int a[300];

int main() {
    scanf("%d", &q);
    while (q--) {
        int n; scanf("%d", &n);
        int p, k;
        rep1(i, 1, n) {
            scanf("%d", &a[i]);
            if (a[i] == 1) k = p = i;
        }
        if (n == 1) {
            puts("YES");
            continue;
        }
        int flag1 = 1, flag2 = 1, last = a[k];
        while (1) {
            p++;
            if (p > n) p = 1;
            if (a[p] - last != 1) {
                flag1 = 0; break;
            }
            last = a[p];
            if (a[p] == n) break;
        }
        last = a[k]; p = k;
        while (1) {
            p--;
            if (!p) p = n;
            if (a[p] - last != 1) {
                flag2 = 0; break;
            }
            last = a[p];
            if (a[p] == n) break;
        }
        if (flag1 || flag2) puts("YES"); else puts("NO");
    }
    return 0;
}

B：

因为面积要相等，显然对棍子排序之后最小匹配最大，次小匹配次大，以此类推。

/* 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 = 500;
int n, a[maxn], t;

int main() {
    scanf("%d", &t);
    while (t--) {
        scanf("%d", &n); n = n * 4;
        for (int i = 1; i <= n; i++) scanf("%d", &a[i]);
        sort(a + 1, a + 1 + n);
        int p = 1, q = n, area = -1, flag = 1;
        while (p < q) {
            if (area == -1) area = a[p] * a[q];
            else if (a[p]*a[q] != area) {
                flag = 0; break;
            }
            if (a[p] != a[p + 1] || a[q] != a[q - 1]) {
                flag = 0; break;
            }
            p += 2, q -= 2;
        }
        if (flag) puts("YES"); else puts("NO");
    }
    return 0;
}

C：

求一遍gcd，答案就是gcd的因数个数。

/* 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 = 4e5 + 10;
int a[maxn], n;

int main() {
    scanf("%d", &n);
    ll gcd = 1, k; scanf("%lld", &k);
    gcd = k;
    for (int i = 1; i < n; i++) {
        ll x; scanf("%lld", &x);
        gcd = __gcd(gcd, x);
    }
    if (gcd == 1) return puts("1"), 0;
    int cnt = 0;
    for (ll i = 1; i < sqrt(gcd); i++)
        if (gcd % i == 0) cnt += 2;
    ll tmp = sqrt(gcd);
    if (tmp * tmp == gcd) cnt++;
    printf("%d
", cnt);
    return 0;
}

D：

/* 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 */

string s, t;

int main() {
    cin >> s >> t;
    int len1 = s.size(), len2 = t.size();
    vector<int>pre(len2), suf(len2);
    int cur = 0;
    // pre存放从前往后扫时字符串t中的字母在字符串s中出现的位置
    for (int i = 0; cur < len2; i++)
        if (s[i] == t[cur])
            pre[cur++] = i;
    // suf存放从后往前扫时字符串t中字母在字符串s中出现的位置
    cur = len2;
    for (int i = len1; i--;) {
        if (s[i] == t[cur - 1])
            suf[--cur] = i;
    }
    // 这样得到的pre数组是字符串t中的每个字母在字符串s中最早出现的下标
    // suf数组同理，得到的是最晚出现的下标
    int ans = max(len1 - 1 - pre.back(), suf[0]);
    for (int i = 0; i + 1 < len2; i++)
        ans = max(ans, suf[i + 1] - pre[i] - 1);
    printf("%d
", ans);
    return 0;
}

E：

给定n个数，每个数x只能变成x-1,x+1或不变。问经过若干次变化后，最多能选出多少个不同的数。

贪心。排序后检查每个数并与一个值last进行比较。若x-1>last，则ans++,last=x-1。其余情况同理。

/* 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 */

int n, ans = 0;
vector<int>a;

int main() {
    scanf("%d", &n);
    for (int i = 1; i <= n; i++) {
        int x; scanf("%d", &x);
        a.pb(x);
    }
    sort(a.begin(), a.end());
    int last = 0;
    for (auto i : a) {
        if (i - 1 > last) ans++, last = i - 1;
        else if (i - 1 == last) ans++, last = i;
        else if (i == last) ans++, last = i + 1;
    }
    printf("%d
", ans);
    return 0;
}

F：