目录
A. Walking Master（贪心）
题面翻译
思路：
代码实现
B. Mex Master（贪心）
题面翻译：
思路：
代码实现
C. Sequence Master（数学）
题面翻译：
思路：
代码实现
A. Walking Master（贪心）
YunQian is standing on an infinite plane with the Cartesian coordinate system on it. In one move, she can move to the diagonally adjacent point on the top right or the adjacent point on the left.
That is, if she is standing on point (x,y), she can either move to point (x+1,y+1) or point (x−1,y)
YunQian initially stands at point (a,b) and wants to move to point (c,d). Find the minimum number of moves she needs to make or declare that it is impossible.
Input
The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of test cases follows.
The first line and only line of each test case contain four integers a, b, c, d(−108≤a,b,c,d≤108).
Output
For each test case, if it is possible to move from point (a,b) to point (c,d), output the minimum number of moves. Otherwise, output −1
Example
input
6
-1 0 -1 2
0 0 4 5
-2 -1 1 1
-3 2 -3 2
2 -1 -1 -1
1 1 0 2
output
4 6 -1 0 3 3
Note
In the first test case, one possible way using 44 moves is (−1,0)→(0,1)→(−1,1)→(0,2)→(−1,2). It can be proven that it is impossible to move from point (−1,0)to point (−1,2) in less than 4 moves.
题面翻译
一个人从（x，y）可以走到（x+1 , y+1）和（x-1，y），问从（a，b）走到（c，d）最少几步
思路：
优先斜着走，b走到d后再横着走到c
走不到的话就输出-1
代码实现
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
struct node{
int x,y;
};
int main(){
int t;
cin>>t;
while(t--){
int x1,y1,x2,y2;
cin>>x1>>y1>>x2>>y2;
if(y2<y1){
cout<<-1<<endl;
continue;
}
int dy=y2-y1;
if(x2-x1>dy){
cout<<-1<<endl;
continue;
}
int x=x1+dy;
cout<<dy+(x-x2)<<endl;
}
return 0;
}
B. Mex Master（贪心）
You are given an array a of length n. The score of a is the MEX†† of [a1+a2,a2+a3,…,an−1+an]. Find the minimum score of a if you are allowed to rearrange elements of a in any order. Note that you are not required to construct the array a that achieves the minimum score.
†† The MEX (minimum excluded) of an array is the smallest non-negative integer that does not belong to the array. For instance:
- The MEX of [2,2,1]is 0, because 0 does not belong to the array.
- The MEX of [3,1,0,1] is 2, because 0 and 1 belong to the array, but 2 does not.
- The MEX of [0,3,1,2] is 4 because 0, 11, 2, and 3 belong to the array, but 4 does not.
Input
The first line contains a single integer t (1≤t≤104) — the number of test cases. The description of test cases follows.
The first line of each test case contains a single integer n (2≤n≤2⋅105).
The second line of each test case contains n integers a1,a2,…,an(0≤ai≤2⋅105).
It is guaranteed that the sum of n over all test cases does not exceed 2⋅1052⋅105.
Output
For each test case, output the minimum score of a after rearranging the elements of a in any order.
Example
input
3
2
0 0
3
0 0 1
8
1 0 0 0 2 0 3 0
output
1 0 1
Note
In the first test case, it is optimal to rearrange a as [0,0], the score of this array is the MEX of [0+0]=[0], which is 1.
In the second test case, it is optimal to rearrange a as [0,1,0], the score of this array is the MEX of [0+1,1+0]=[1,1], which is 0.
题面翻译：
给你一个长为n的数列，你对这个数列进行排序，然后构造一个长为n-1新数列，每一项由排序后的相邻两项相加而成，求新数列中不存在的最小非负整数
思路：
如果0的个数少于（n+1）/2，可以构造一个不含0的数列，最小整数为0
0和1的个数小于n，可以把0放在一边，1放在另一半，构造出的数列不含1，最小整数为1,
数列全由0,1组成，用0把1隔开，最小不存在的数为2
代码实现
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int cnt[200005];
int main(){
int t;
cin>>t;
while(t--){
int n;
cin>>n;
memset(cnt,0,sizeof(cnt));
for(int i=1;i<=n;i++){
int a;
cin>>a;
cnt[a]++;
}
if(cnt[0]<=(n+1)/2){
cout<<0<<endl;
continue;
}
if(cnt[1]+cnt[0]<n){
cout<<1<<endl;
continue;
}
if(cnt[1]==0){
cout<<1<<endl;
continue;
}
cout<<2<<endl;
}
return 0;
}
C. Sequence Master（数学）
For some positive integer m, YunQian considers an array q of 2m (possibly negative) integers good, if and only if for every possible subsequence of q that has length m, the product of the m elements in the subsequence is equal to the sum of the m elements that are not in the subsequence. Formally, let U={1,2,…,2m} For all sets S⊆U such that |S|=m|=, ∏i∈Sqi=∑i∈U∖Sqi∏.
Define the distance between two arrays a and bboth of length k\ to be ∑i=1k|ai−bi|.
You are given a positive integer n and an array p of 2n integers.
Find the minimum distance between p and q over all good arrays q of length 2n. It can be shown for all positive integers n, at least one good array exists. Note that you are not required to construct the array q that achieves this minimum distance.
Input
The first line contains a single integer t (1≤t≤104 — the number of test cases. The description of test cases follows.
The first line of each test case contains a single integer n (1≤n≤2⋅105).
The second line of each test case contains 2n integers p1,p2,…,p2n (|pi|≤109).
It is guaranteed that the sum of n over all test cases does not exceed 2⋅105.
Output
For each test case, output the minimum distance between p and a good q.
Example
input
4
1
6 9
2
1 2 2 1
2
-2 -2 2 2
4
-3 -2 -1 0 1 2 3 4
output
Copy
3 2 5 13
Note
In the first test case, it is optimal to let q=[6,6].
In the second test case, it is optimal to let q=[2,2,2,2].
题面翻译：
给你一个长为2*n的数列，找一个长为2*n的数列满足任意n项的和等于另外n项的积，使得这两个数列每一项之差的绝对值的和最小
思路：
md纯一个数学题
n=1的时候就是两个数的差值
n=2的时候，三种情况，全是2，全是0，3个-1和一个2，全部讨论取最小值即可
n%2=0的时候，要么全是0，要么是2n-1个-1和一个n，讨论文章来源：https://www.toymoban.com/news/detail-403303.html
n%2=1的时候，全是0文章来源地址https://www.toymoban.com/news/detail-403303.html
代码实现
#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
int a[400005];
int main(){
int t;
cin>>t;
while(t--){
ll n,ans=LLONG_MAX;
cin>>n;
for(int i=0;i<2*n;i++){
cin>>a[i];
}
if(n==1){
ans=abs(a[1]-a[0]);
}
if(n==2){
ll sum=0;
for(int i=0;i<2*n;i++){
sum+=abs(a[i]-2);
}
ans=sum;
}
if(n%2==0){
ll sum=0;
for(int i=0;i<2*n;i++){
sum+=abs(a[i]+1);
}
for(int i=0;i<2*n;i++){
ans=min(ans,sum-abs(a[i]+1)+abs(a[i]-n));
}
}
ll sum=0;
for(int i=0;i<2*n;i++){
sum+=abs(a[i]);
}
ans=min(ans,sum);
cout<<ans<<endl;
}
return 0;
}
到了这里，关于Codeforces Round 858 (Div. 2)题解的文章就介绍完了。如果您还想了解更多内容，请在右上角搜索TOY模板网以前的文章或继续浏览下面的相关文章，希望大家以后多多支持TOY模板网！
