链接：
https://codeforces.com/contest/1265/problem/D
题意：
An integer sequence is called beautiful if the difference between any two consecutive numbers is equal to 1. More formally, a sequence s1,s2,…,sn is beautiful if |si−si+1|=1 for all 1≤i≤n−1.
Trans has a numbers 0, b numbers 1, c numbers 2 and d numbers 3. He wants to construct a beautiful sequence using all of these a+b+c+d numbers.
However, it turns out to be a non-trivial task, and Trans was not able to do it. Could you please help Trans?
思路：
考虑0只能和1，3只能和2，贪心的以01， 23去放，这样0必须小于等于1， 2和3同理，再往中间插21，如果剩下的2和1相差超过1就不能满足条件。
多一个可以插在开头或者结尾。还要考虑0和1为0，2和3为0的特殊情况。
好像暴力枚举也可以。。
代码：
#include<bits/stdc++.h>
using namespace std;
int a, b, c, d;
int main()
{
cin >> a >> b >> c >> d;
if (((d != 0 || c != 0) && a > b) || ((a != 0 || b != 0) && d > c))
{
puts("NO\n");
return 0;
}
if (a == 0 && b == 0)
{
if (abs(d-c) > 1)
{
puts("NO\n");
return 0;
}
else
{
string res = "";
for (int i = 1;i <= min(c, d);i++)
res += "23";
if (c > d)
res += "2";
else if (c < d)
res = "3"+res;
cout << "YES" << endl;
for (int i = 0;i < res.length();i++)
cout << res[i] << ' ';
cout << endl;
return 0;
}
}
if (c == 0 && d == 0)
{
if (abs(a-b) > 1)
{
puts("NO\n");
return 0;
}
else
{
string res = "";
for (int i = 1;i <= min(a, b);i++)
res += "01";
if (a > b)
res += "0";
else if (a < b)
res = "1"+res;
cout << "YES" << endl;
for (int i = 0;i < res.length();i++)
cout << res[i] << ' ' ;
cout << endl;
return 0;
}
}
int l = b-a, r = c-d;
if (abs(l-r) > 1)
{
puts("NO\n");
return 0;
}
string res = "";
if (l-r == 1)
res += "1";
for (int i = 1;i <= a;i++)
res += "01";
for (int i = 1;i <= min(l, r);i++)
res += "21";
for (int i = 1;i <= d;i++)
res += "23";
if (r-l == 1)
res += "2";
cout << "YES" << endl;
for (int i = 0;i < res.length();i++)
cout << res[i] << ' ';
cout << endl;
return 0;
}