Codeforces Round #956 (Div. 2) and ByteRace 2024（A~D题解）
这次比赛也是比较吃亏的，做题顺序出错了，先做的第三个，错在第三个数据点之后，才做的第二个（因为当时有个地方没检查出来）所以这次比赛还是一如既往地打拉了
那么就来发一下题解吧
A. Array Divisibility
题意：对于1<=k<=n，对于每个k其倍数下标之和一定为k的倍数
思路：直接从1赋值到n就行，也是水题一个
#include<bits/stdc++.h>
using namespace std;
#define int long long
int t;
int n;
signed main()
{cin>>t;while(t--){cin>>n;for(int i=1;i<=n;i++){cout<<i<<" ";}cout<<"\n";}return 0;
}
B. Corner Twist
题意：就是给你两个数组，问你两个数组能否按照题上所说的方法相互转换得到
思路：将整个大矩阵拆成2*2的小矩阵，然后每次只要让左上角那个和下面的变成一样就可以，然后我们最后原本只需要检查最后一列和最后一行是否相同就可以（ps：我写的是逐一比较，因为比较好写）（错了一次是因为比较的时候内层循环写成n了）
#include<bits/stdc++.h>
using namespace std;
#define int long long
int t;
int n,m;
char a[505][505];
char b[505][505];
signed main()
{cin>>t;while(t--){cin>>n>>m;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>a[i][j];}}for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){cin>>b[i][j];}}for(int i=1;i<=n-1;i++){for(int j=1;j<=m-1;j++){int cha = ((b[i][j]-'0')-(a[i][j]-'0')+3)%3;if(cha==1){a[i][j]=(a[i][j]+1)%3+'0';a[i+1][j+1]=(a[i+1][j+1]+1)%3+'0';a[i+1][j]=(a[i+1][j]+2)%3+'0';a[i][j+1]=(a[i][j+1]+2)%3+'0';}else if(cha==2){a[i][j]=(a[i][j]+2)%3+'0';a[i+1][j+1]=(a[i+1][j+1]+2)%3+'0';a[i+1][j]=(a[i+1][j]+1)%3+'0';a[i][j+1]=(a[i][j+1]+1)%3+'0';}}}int flag=1;for(int i=1;i<=n;i++){for(int j=1;j<=m;j++){if(a[i][j]==b[i][j]){continue;}elseflag=0;}}if(flag==0){cout<<"NO"<<"\n";}else{cout<<"YES"<<"\n";}}return 0;
}C. Have Your Cake and Eat It Too
题意：就说有一个蛋糕 ，被分成了许多块，然后三个人对每部分的蛋糕都有一个自己的价值，但是所有块的价值总和是一定的，然后问你如何划分这个区间，才能满足每个区间都大于(tot+2)/3
思路：对六种情况分别贪心即可，先让两边的取到比sum大的位置，然后再看中间的是否比sum大，是的话就直接输出，如果都不满足，最后就只能输出-1了
#include<bits/stdc++.h>
using namespace std;
#define int long long
signed main()
{int t;cin>>t;while(t--){int n;cin >> n;int a[n+1], b[n+1], c[n+1];int sum = 0;for(int i = 1;i<=n;i++){cin >> a[i];sum += a[i];}sum = (sum+2)/3;//题中说了上限除法 for(int i = 1;i<=n;i++){cin >> b[i];}for(int i = 1;i<=n;i++){cin >> c[i];}vector<int> p1(n+5), p2(n+5), p3(n+5);//正序前缀和 vector<int> s1(n+5), s2(n+5), s3(n+5);//倒序前缀和 for(int i = 1; i <= n; i++){p1[i] = p1[i-1] + a[i];p2[i] = p2[i-1] + b[i];p3[i] = p3[i-1] + c[i];}for(int i = n; i >= 1; i--){s1[i] = s1[i+1] + a[i];s2[i] = s2[i+1] + b[i];s3[i] = s3[i+1] + c[i];}//a b c int i = 1, j = n;while(p1[i-1] < sum && i <= n){i++;}while(s3[j+1] < sum && j >= 1){j--;}if(i <= j && p2[j]-p2[i-1] >= sum){cout << 1 << ' ' << i-1 << ' ' << i << ' ' << j << ' ' << j+1 << ' ' << n << endl;continue;}// a c bi = 1, j = n;while(p1[i-1] < sum && i <= n){i++;}while(s2[j+1] < sum && j >= 1){j--;}if(i <= j && p3[j]-p3[i-1] >= sum){cout << 1 << ' ' << i-1 << ' ' << j+1 << ' ' << n << ' ' << i << ' ' << j << endl;continue;}// b c ai = 1, j = n;while(p2[i-1] < sum && i <= n){i++;}while(s1[j+1] < sum && j >= 1){j--;}if(i <= j && p3[j]-p3[i-1] >= sum){cout << j+1 << ' ' << n << ' ' << 1 << ' ' << i-1 << ' ' << i << ' ' << j << endl;continue;}// b a ci = 1, j = n;while(p2[i-1] < sum && i <= n){i++;}while(s3[j+1] < sum && j >= 1){j--;}if(i <= j && p1[j]-p1[i-1] >= sum){cout << i << ' ' << j << ' ' << 1 << ' ' << i-1 << ' ' << j+1 << ' ' << n << endl;continue;}// c a bi = 1, j = n;while(p3[i-1] < sum && i <= n){i++;}while(s2[j+1] < sum && j >= 1){j--;}if(i <= j && p1[j]-p1[i-1] >= sum){cout << i << ' ' << j << ' ' << j+1 << ' ' << n << ' ' << 1 << ' ' << i-1 << endl;continue;}// c b ai = 1, j = n;while(p3[i-1] < sum && i <= n){i++;}while(s1[j+1] < sum && j >= 1){j--;}if(i <= j && p2[j]-p2[i-1] >= sum){cout << j+1 << ' ' << n << ' ' << i << ' ' << j << ' ' << 1 << ' ' << i-1 << endl;continue;}cout << -1 << endl;}return 0;
}D. Swap Dilemma
题意：就是说给你两个数组，然后每次再a数组选两个坐标，b数组选两个坐标，然后各自再各自的数组交换，然后问你最后两个数组能不能变成一样的
思路：这题我想到了两种做法
逆序对法
（1）逆序对的方法，众所周知，在大学有一门神奇的科目叫做线性代数，线性代数里面讲过一个东西叫做逆序对，只有逆序对的个数为同一奇偶性，才有可能相同，因为a，b数组每次都要变换一次，所以他们的奇偶性一定是都会在每一次变化，所以我们需要统计奇偶性，然后来判断，当然了，在之前还需要判断元素种类是否相同，如果个数不同一定为no
#include<bits/stdc++.h>
using namespace std;
#define int long long
int mergeSort(vector<int> &nums, int left, int right)
{if (left >= right) {return 0;}int mid = left + (right - left) / 2;int count = mergeSort(nums, left, mid) + mergeSort(nums, mid + 1, right);vector<int> tmp(right - left + 1);int i = left, j = mid + 1, k = 0;while (i <= mid && j <= right) {if (nums[i] <= nums[j]) {tmp[k++] = nums[i++];} else {tmp[k++] = nums[j++];count += mid - i + 1; // 计算逆序数}}while (i <= mid) {tmp[k++] = nums[i++];}while (j <= right) {tmp[k++] = nums[j++];}for (int p = 0; p < tmp.size(); ++p) {nums[left + p] = tmp[p];}return count;
}int solve(vector<int> &nums)
{if (nums.size() <= 1) {return 0;}return mergeSort(nums, 0, nums.size() - 1);
}void solve()
{map<int,int> mp;int n;cin >> n;vector<int> a(n+2);vector<int> b(n+2);for (int i=1;i<=n;i++) cin >> a[i];for (int i=1;i<=n;i++){cin >> b[i];mp[b[i]]=i;}for(int i=1;i<=n;i++){if(mp.count(a[i])==0){cout<<"NO\n";return ;}}int ans1=solve(a),ans2=solve(b);if (ans1%2 ==ans2%2) cout << "YES\n";else cout << "NO\n";
}
signed main()
{int t;cin >> t;while (t--) solve();return 0;
}交换次数法
（2）那么来讲另一种比较简单的方法，交换次数来判断，因为题目上所说每次两个数组都要交换，那么我们就只交换一个，然后统计变成另一个的次数为多少，是偶数就是yes是奇数就是no
当然了，在之前也是需要判断种类是否相同的
#include <bits/stdc++.h>
using namespace std;
#define int long long
int a[200005], b[200005];
map<int, int> mp;
void solve()
{mp.clear();int n;cin >> n;for (int i=1;i<=n;i++) cin >> a[i];for (int i=1;i<=n;i++){cin >> b[i];mp[b[i]]=i;}int ans = 0;for (int i=1;i<=n;i++){if (b[i] == a[i]) continue;if (mp.count(a[i]) == 0){cout << "NO\n";return;}int p=mp[a[i]];swap(b[i],b[p]);mp[b[i]]=i;mp[b[p]]=p;ans+=1; }if (ans%2 == 0) cout << "YES\n";else cout << "NO\n";
}
signed main()
{int t;cin >> t;while (t--) solve();return 0;
}