Codeforces Round #740 Div. 2

题目跳转链接

A. Simply Strange Sort

题意

定义一个函数\(f_{i}\) : 如果\(a_i \ge a_{i+1}\) swap(\(a_i\) \(a_{i+1}\))

定义一个操作：

第奇数次是 执行 \(f_1\) \(f_3\)... \(f_{n-2}\)

第偶数次是 执行 \(f_2\) \(f_4\)... \(f_{n-1}\)

求出进行多少次操作可以使原数组变得有序

思路

模拟上述操作即可 判断符合条件也可以用 is_sorted函数

AC_CODE

#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define mk make_pair
#define debug(x) cout<<#x" ----> "<<x<<endl
#define rep(i, b, s) for(int i = (b); i <= (s); ++i)
#define pre(i, b, s) for(int i = (b); i >= (s); --i)

//#define int long long
#define endl '\n'
#define ios ios::sync_with_stdio(false); cin.tie(0), cout.tie(0)
#define all(v) (v).begin(),(v).end()

using namespace std;

typedef unsigned long long ULL;
typedef pair<int, int> PII ;
typedef pair<double, double> PDD ;
typedef long long LL;
const int INF = INT_MAX;
const int mod = 1e9 + 7;
const double eps = 1e-10;
const double pi = acos(-1.0);

int lowbit(int x){return x&-x;}
int gcd(int a, int b) {return b ? gcd(b, a%b) : a;}
LL ksm(LL a, LL b) {if (b == 0) return 1; LL ns = ksm(a, b >> 1); ns = ns * ns % mod; if (b & 1) ns = ns * a % mod; return ns;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
template < typename T >
inline void read(T &x)
{
    x = 0; bool f = 0; char ch = getchar();
    while(!isdigit(ch)){f ^= !(ch ^ 45);ch=getchar();}
    while(isdigit(ch)) x= (x<<1)+(x<<3)+(ch&15),ch=getchar();
    x = f ? -x : x;
}

void solve() {
    int n; read(n);
    vector<int> a(n + 1), b(n + 1);
    for(int i = 1; i <= n; i ++ ) {
        read(a[i]);
        b[i] = i;
    }
    int cnt = 0;
    for(int i = 1; i <= n; i ++ ) {
        if(a == b) {
            break;
        }

        //// if(is_sorted(all(a))) break;

        if(i & 1) {
            for(int j = 1; j <= n - 2; j += 2)
                if(a[j] > a[j + 1]) swap(a[j], a[j + 1]);
        }
        else {
            for(int j = 2; j <= n - 1; j += 2)
                if(a[j] > a[j + 1]) swap(a[j], a[j + 1]);
        }
        cnt ++;

    }
    printf("%d\n", cnt);

}

signed main()
{
    int T = 1;  scanf("%d",&T);
    while(T -- ) {
        solve();
    }

    return 0;
}

B. Charmed by the Game

题意

题目背景: 乒乓球比赛

规定乒乓球发球为两方选手交替发球

如果本场A选手发球

- A赢了 被称为 A hold

- B赢了 被称为 B break

如果本场B选手发球

- B赢了 被称为 B hold

- A赢了 被称为 A break

给出 A 和 B 赢得次数

求出有多少种 break

思路

我们可以发现赢得场次是固定的, 因此我们只需要枚举A在发球场次赢得个数 就可以的到全部的break

reason 我们枚举了A在发球场次赢得个数 因此我们可以得到另外三种情况,就可以求出所有的break

注意: 第一场发球可能是A 也可能是B 因此 我们只需要求出A先发球 所有的break

然后用a+b-i 就是B先发球的break

AC_CODE

#include <bits/stdc++.h>
#define x first
#define y second
#define pb push_back
#define mk make_pair
#define debug(x) cout<<#x" ----> "<<x<<endl
#define rep(i, b, s) for(int i = (b); i <= (s); ++i)
#define pre(i, b, s) for(int i = (b); i >= (s); --i)

//#define int long long
#define endl '\n'
#define ios ios::sync_with_stdio(false); cin.tie(0), cout.tie(0)
#define all(v) (v).begin(),(v).end()

using namespace std;

typedef unsigned long long ULL;
typedef pair<int, int> PII ;
typedef pair<double, double> PDD ;
typedef long long LL;
const int INF = INT_MAX;
const int mod = 1e9 + 7;
const double eps = 1e-10;
const double pi = acos(-1.0);

int lowbit(int x){return x&-x;}
int gcd(int a, int b) {return b ? gcd(b, a%b) : a;}
LL ksm(LL a, LL b) {if (b == 0) return 1; LL ns = ksm(a, b >> 1); ns = ns * ns % mod; if (b & 1) ns = ns * a % mod; return ns;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
template < typename T >
inline void read(T &x)
{
    x = 0; bool f = 0; char ch = getchar();
    while(!isdigit(ch)){f ^= !(ch ^ 45);ch=getchar();}
    while(isdigit(ch)) x= (x<<1)+(x<<3)+(ch&15),ch=getchar();
    x = f ? -x : x;
}
const int N = 2e5 + 10;
bool num[N];
int cnt, mid, px;
void solve() {
    int a, b;
    read(a); read(b);
    px = a + b;
    for(int i = 0; i <= px + 5; i ++ ) num[i] = false;
    if(a < b) swap(a, b);
    mid = (px + 1) / 2;
    for(int i = mid; i >= mid - b; i -- ) { //i是A在他发球的场次赢得次数
        cnt = mid - i + a - i;
        num[cnt] = true;
        num[px - cnt] = true;
    }
    vector<int> ans;
    for(int i = 0; i <= px; i ++ )
        if(num[i])
            ans.pb(i);
    printf("%d\n", ans.size());
    for(int p : ans) printf("%d ", p);
    puts("");
}

signed main()
{
    int T = 1;  scanf("%d",&T);
    while(T -- ) {
        solve();
    }

    return 0;
}

C - Deep Down Below

题意

略

思路

按照通过每个关卡所需要的最小力量排序, 然后按照这个顺序枚举即可

AC_CODE

#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define mk make_pair
#define debug(x) cout<<#x" ----> "<<x<<endl
#define debug2(x) cout<<#x" ----> "<<x<< ' '
#define rep(i, b, s) for(int i = (b); i <= (s); ++i)
#define pre(i, b, s) for(int i = (b); i >= (s); --i)

//#define int long long
#define endl '\n'
#define ios ios::sync_with_stdio(false); cin.tie(0), cout.tie(0)
#define all(v) (v).begin(),(v).end()

using namespace std;

typedef unsigned long long ULL;
typedef pair<int, int> PII ;
typedef pair<double, double> PDD ;
typedef long long LL;
const int INF = INT_MAX;
const int mod = 1e9 + 7;
const double eps = 1e-10;
const double pi = acos(-1.0);

int lowbit(int x){return x&-x;}
int gcd(int a, int b) {return b ? gcd(b, a%b) : a;}
LL ksm(LL a, LL b) {if (b == 0) return 1; LL ns = ksm(a, b >> 1); ns = ns * ns % mod; if (b & 1) ns = ns * a % mod; return ns;}
LL lcm(LL a, LL b) {return a / gcd(a, b) * b;}
template < typename T >
inline void read(T &x)
{
    x = 0; bool f = 0; char ch = getchar();
    while(!isdigit(ch)){f ^= !(ch ^ 45);ch=getchar();}
    while(isdigit(ch)) x= (x<<1)+(x<<3)+(ch&15),ch=getchar();
    x = f ? -x : x;
}
const int N = 1e5 + 10;
vector<int> a[N];
PII num[N];
int need[N];
int t, n, rr;

inline void solve() {
    scanf("%d", &t);

    rep(i, 0, t - 1) {
        a[i].clear();
        scanf("%d", &n);
        rep(j, 1, n) {
            scanf("%d", &rr);
            a[i].pb(rr);
        }
        int res = 0;
        int st = 0;
        int cnt = 0;
        rep(j, 0, n - 1) {
            if(a[i][j] >= res) {
                st = a[i][j] + 1 - cnt;
                res = a[i][j] + 2;
            }
            else {
                res ++;
            }
            cnt ++;
        }
        need[i] = st;
        num[i].fi = st;
        num[i].se = i;
    }
    sort(num, num + t);

    int st = 0, res = 0, cnt = 0;
    for(int i = 0; i < t; i ++ ) {
        int idx = num[i].se;
        for(int j = 0; j < a[idx].size(); j ++ ) {
            if(a[idx][j] >= res) {
                st = a[idx][j] + 1 - cnt;
                res = a[idx][j] + 2;
            }
            else {
                res ++;
            }
            cnt ++;
        }
    }
    printf("%d\n", st);

}

signed main()
{
    int T = 1;  scanf("%d",&T);
    while(T -- ) {
        solve();
    }

    return 0;
}

D1. Up the Strip (simplified version)

整除分块

我们枚举 \(l\) 因此\(i/l\) 可以由\(i\) 除以 (l ---> i/(i/l)) 这个区间内的数字得到

这就是整除分块, 因此我们可以通过这个转移方程求出通过除法转移过来的

减法通过求后缀和可以O(1) 求出

AC_CODE

#include <bits/stdc++.h>
using namespace std;
#define int long long
int mod;
int x[1010000];
int getmax(int n,int l)
{
    return n/(n/l);
}
signed main()
{
    int n;
    scanf("%lld%lld",&n,&mod);
    x[n] = 1;
    int sum = 0;
    for(int i = n;i>=2;i--)
    {
        x[i] = (x[i] + sum) % mod;
        for(int l = 2,r;l<=i;l=r+1)
        {
            r = getmax(i,l);
            x[i/l] = (x[i/l] + (r-l+1) * x[i]) % mod;
        }
        sum = (sum + x[i]) %mod;
    }
    printf("%lld",(x[1] + sum)%mod);
    return 0;
}

D2. Up the Strip

枚举倍数

对于每个\(i\) 只有\([i \times j, (i+1) \times j)\)区间内的数字才可以通过除以\(j\) 得到 （j是枚举出的每个被除数0）

对于减法 我们加上后缀和即可

AC_CODE

#include <bits/stdc++.h>
#define LL long long

using namespace std;

const int N = 4e6 + 10;
LL n, m;
LL dp[N];
LL s[N];

LL sum(int l, int r) {
    return (s[l] - s[r + 1] + m) % m;
}

int main() {
    scanf("%lld%lld", &n, &m);
    dp[n] = 1;
    s[n] = 1;
    for(int i = n - 1; i; i --) {
        dp[i] = s[i + 1];
        for(int j = 2; j <= n / i; j ++ ) {
            int l = i * j, r = min(n, (LL)j * (i + 1) - 1);
            dp[i] = dp[i] + sum(l, r);
            dp[i] %= m;
        }
        s[i] = s[i + 1] + dp[i];
        s[i] %= m;
    }
    printf("%lld", (dp[1] + m) % m);

    return 0;
}