Largest Beautiful Number CodeForces - 946E (贪心)

大意: 定义一个好数为位数为偶数, 且各位数字重排后可以为回文, 对于每个询问, 求小于$x$的最大好数.

 

假设$x$有$n$位, 若$n$为奇数, 答案显然为$n-1$个9. 若为偶数, 我们想让答案尽量大, 那么就要尽量调整$x$的低位数, 从低位到高位遍历, 假设当前处理到第$i$位, 对于判断重排后回文可以用一个长为10的二进制数来判断, 假设$[i+1,n]$状态为$t$, 那么问题就转化为求最大的小于$t$且二进制状态为$s$的数, 贪心填数即可, 若对于每一位都无解答案就为$n-2$个9.

第一次写这种涉及数位的贪心模拟, 感觉还是挺烦的, 进位要多注意, 贪心时的因为要多次进行可行性判断, 一定要写成函数.

#include <iostream>
#include <sstream>
#include <algorithm>
#include <cstdio>
#include <math.h>
#include <set>
#include <map>
#include <queue>
#include <string>
#include <string.h>
#include <bitset>
#define REP(i,a,n) for(int i=a;i<=n;++i)
#define PER(i,a,n) for(int i=n;i>=a;--i)
#define hr putchar(10)
#define pb push_back
#define lc (o<<1)
#define rc (lc|1)
#define mid ((l+r)>>1)
#define ls lc,l,mid
#define rs rc,mid+1,r
#define x first
#define y second
#define io std::ios::sync_with_stdio(false)
#define endl '\n'
#define DB(a) ({REP(__i,1,n) cout<<a[__i]<<' ';hr;})
using namespace std;
typedef long long ll;
typedef pair<int,int> pii;
const int P = 1e9+7, INF = 0x3f3f3f3f;
ll gcd(ll a,ll b) {return b?gcd(b,a%b):a;}
ll qpow(ll a,ll n) {ll r=1%P;for (a%=P;n;a=a*a%P,n>>=1)if(n&1)r=r*a%P;return r;}
ll inv(ll x){return x<=1?1:inv(P%x)*(P-P/x)%P;}
inline int rd() {int x=0;char p=getchar();while(p<'0'||p>'9')p=getchar();while(p>='0'&&p<='9')x=x*10+p-'0',p=getchar();return x;}
//head



#ifdef ONLINE_JUDGE
const int N = 1e6+10;
#else
const int N = 111;
#endif

char s[N];
int n, f[N], cnt[N], num[N];
int pos[N], p[15];

int chk(int i, int sta, int flag) {
	if (i>n) return 1;
	int sz = 0;
	REP(i,0,9) if (sta>>i&1) p[++sz]=i;
	if (sz>n-i+1||(sz&1)!=(n-i+1&1)) return 0;
	if (flag) return 1;
	if (i+pos[i]-1+sz>n) return 0;
	if (i+pos[i]-1+sz==n) {
		int ok = 0;
		REP(j,i+pos[i],n) { 
			if (s[j]<p[j-i-pos[i]+1]) return 0;
			if (s[j]>p[j-i-pos[i]+1]) return 1;
		}
		return 0;
	}
	return 1;
}

void work() {
	scanf("%s", s+1);
	n = strlen(s+1);
	if (n&1) { 
		REP(i,1,n-1) putchar('9');hr;
		return;
	}
	REP(i,1,n) s[i]-='0';
	REP(i,1,n) f[i]=f[i-1]^(1<<s[i]);
	pos[n+1]=0;
	PER(i,1,n) pos[i]=s[i]?0:pos[i+1]+1;
	PER(i,1,n) {
		int sta = f[i-1];
		if (!chk(i,sta,0)) continue;
		int flag = 0;
		REP(j,i,n-1) {
			if (!flag) {
				if (chk(j+1,sta^1<<s[j],flag)) {
					sta ^= s[j];
					continue;
				}
				flag = 1;
				PER(k,0,s[j]-1) if (chk(j+1,sta^1<<k,flag)) {
					s[j] = k, sta ^= 1<<k;
					break;
				}
			}
			else {
				PER(k,0,9) if (chk(j+1,sta^1<<k,1)) {
					s[j] = k, sta ^= 1<<k;
				}
			}
		}
		REP(j,0,9) if (sta>>j&1) s[n]=j;
		if (!s[1]) break;
		REP(j,1,n) printf("%d", s[j]);hr;
		return;
	}
	REP(i,1,n-2) putchar('9');hr;
}

int main() {
	int t;
	scanf("%d", &t);
	while (t--) work();
}