Educational Codeforces Round 39: E. Largest Beautiful Number

E. Largest Beautiful Numbertime limit per test 1 secondmemory limit per test 256 megabytesinput standard inputoutput standard outputYes, that's another problem with definition of "beautiful" numbers.Let's call a positive integer x beautiful if its decimal representation without leading zeroes contains even number of digits, and there exists a permutation of this representation which is palindromic. For example, 4242 is a beautiful number, since it contains 4 digits, and there exists a palindromic permutation 2442.Given a positive integer s, find the largest beautiful number which is less than s.InputThe first line contains one integer t (1 ≤ t ≤ 105) — the number of testcases you have to solve.Then t lines follow, each representing one testcase and containing one string which is the decimal representation of number s. It is guaranteed that this string has even length, contains no leading zeroes, and there exists at least one beautiful number less than s.The sum of lengths of s over all testcases doesn't exceed 2·105.OutputFor each testcase print one line containing the largest beautiful number which is less than s (it is guaranteed that the answer exists).ExampleinputCopy

4
89
88
1000
28923845

output

88
77
99
28923839

题意：如果一个数满足①长度为偶数；②数字0~9都出现偶数次，那么这个数就是美丽的，给你一个n，求出小于n的最大的美丽数字，输入保证有解，答案不能有前导0

从最低位到最高位暴力就行了，对于当前第k位，如果将这个位子上的数字-1，那么后面所有数字都可以随意
考虑贪心，前面所有出现奇数次的数字从小到大依次从低位填到高位，然后中间全部填9（可以证明如果还有空隙，那么空隙也一定刚好有偶数个）填不下去就说明当前不行，然后这个位置上的数字-2继续判直到减到0为止
中间有解就直接输出，还有注意已经暴力到最高位了并且减到了0还不行，就直接输出len-2个9（len为数字n长度）

#include<stdio.h>
#include<string.h>
#include<algorithm>
using namespace std;
char str[200005];
int flag[12], temp[12];
int main(void)
{
int T, n, i, j, k, cnt;
scanf("%d", &T);
while(T--)
{
scanf("%s", str+1);
memset(flag, 0, sizeof(flag));
n = strlen(str+1);
if(n%2==1)
{
for(i=1;i<=n-1;i++)
printf("9");
printf("\n");
continue;
}
for(i=1;i<=n;i++)
flag[str[i]-'0'] ^= 1;
cnt = 0;
for(i=n;i>=1;i--)
{
if(str[i]=='0')
{
flag[str[i]-'0'] ^= 1;
continue;
}
for(k=1;str[i]-k>='0';k++)
{
flag[str[i]-'0'] ^= 1;
flag[str[i]-k-'0'] ^= 1;
if(i==1 && str[i]-k=='0')
{
for(i=1;i<=n-2;i++)
printf("9");
puts("");
goto loop;
}
cnt = 0;
for(j=0;j<=9;j++)
{
if(flag[j])
temp[++cnt] = j;
}
if(cnt<=n-i)
{
for(j=1;j<=cnt;j++)
str[n-j+1] = temp[j]+'0';
for(j=n-cnt;j>=i+1;j--)
str[j] = '9';
str[i] = str[i]-k;
puts(str+1);
goto loop;
}
flag[str[i]-'0'] ^= 1;
flag[str[i]-k-'0'] ^= 1;
}
flag[str[i]-'0'] ^= 1;
}
loop:;
}
}
/*
1
9000
*/