Codeforces Round #740 Div. 2
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;
}
