Codeforces Round #562 (div.2)

跟之前AtCoder BC 128同一晚上的题，没有打。

这场题面很短，好评。

题目链接：https://codeforces.com/contest/1169

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A：

题目又臭又长，结果是个弱智模拟。

/* 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, a, x, b, y;

int main() {
    cin >> n >> a >> x >> b >> y;
    while ((a != x) && (b != y)) {
        a = a == n ? 1 : a + 1;
        b = b == 1 ? n : b - 1;
        if (a == b) return puts("YES"), 0;
    }
    puts("NO");
    return 0;
}

B：

也很弱智。

/* 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 = 3e5 + 10;
int n, m, cnt[maxn];
vector<pair<int, int>>v;

int check(int x) {
    rep0(i, 0, maxn) cnt[i] = 0;
    int s = 0;
    rep0(i, 0, m)
    if (v[i].first != x && v[i].second != x) {
        cnt[v[i].first]++, cnt[v[i].second]++;
        s++;
    }
    rep1(i, 1, n)
    if (cnt[i] == s)
        return 1;
    return 0;
}

int main() {
    scanf("%d%d", &n, &m);
    rep1(i, 1, m) {
        int x, y; scanf("%d%d", &x, &y);
        v.pb(mp(x, y));
    }
    if (check((*v.begin()).first) || check((*v.begin()).second))
        puts("YES");
    else puts("NO");
    return 0;
}

C：

给定n、m和n个数，每个数的范围是[0,m)。有且仅有一种操作：每次操作可以选定任意k个数，对于每个数a[i]，a[i]=(a[i]+1)%m。问最少需要几次操作，使得原数列变为不下降数列。

数据范围3e5，一看就是二分题。

/* 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 = 3e5 + 10;
int n, m, a[maxn];

int main() {
    scanf("%d%d", &n, &m);
    rep0(i, 0, n) scanf("%d", &a[i]);
    int l = -1, r = m;
    while (l < r - 1) {
        int mid = l + r >> 1, prev = 0, flag = 1;
        rep0(i, 0, n) {
            int lpos = a[i], rpos = a[i] + mid;
            if ((lpos <= prev && prev <= rpos) || (lpos <= prev + m && prev + m <= rpos))
                continue;
            if (a[i] > prev) prev = a[i];
            else {
                flag = 0;
                break;
            }
        }
        if (flag) r = mid; else l = mid;
    }
    printf("%d
", r);
    return 0;
}

D：

给定一个长度为n的01串s，求有多少对l、r (1≤l≤r≤n)，存在至少一对x、k，满足1≤x,k≤n，l≤x<x+2k≤r，且s[x]=s[x+k]=s[x+2k]。

翻译一下就是有多少个区间，区间内至少存在一个三元组(a,b,c)，a==b==c且间隔相等。

暴力，枚就完事了。

/* 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 = 3e5 + 10;
int a[maxn];
string s;
ll ans = 0;

int main() {
    cin >> s;
    int len = s.size();
    rep0(i, 0, maxn) a[i] = int_inf;
    rep0(i, 0, len) {
        for (int j = 1; i + 2 * j < len; j++)
            if (s[i] == s[i + j] && s[i] == s[i + 2 * j]) {
                a[i] = i + 2 * j;
                break;
            }
    }
    for (int i = len - 1; i >= 0; i--) a[i] = min(a[i], a[i + 1]);
    rep0(i, 0, len)
    if (a[i] != int_inf) ans += len - a[i];
    printf("%lld
", ans);
    return 0;
}

E：

待补。