原题链接
考察:数位dp
思路:
  求回文数字的个数.
  dfs参数:
1.pos 枚举到第几位
2.len 回文数的长度.
  其实不需要变量记录是否合法,不合法的不取搜就行了.

Code

#include <iostream>
#include <cstring>
using namespace std;
typedef long long LL;
const int N = 20;
LL l,r,f[N][N];
int a[N],nums[N];
LL dfs(int pos,int len,bool limit,bool lead)
{
	if(!pos) return 1;
	if(!limit&&!lead&&~f[pos][len]) return f[pos][len];
	int up = limit?a[pos]:9;
	LL res = 0;
	for(int i=0;i<=up;i++)
	{
		if(!i&&lead) res+=dfs(pos-1,len,limit&&i==up,lead&&!i);
		else{
			nums[pos] = i;
			if(lead) res+=dfs(pos-1,pos,limit&&i==up,lead&&!i);
			else if(pos>=len/2+1) res+=dfs(pos-1,len,limit&&i==up,lead&&!i);
			else if(i==nums[len+1-pos]) res+=dfs(pos-1,len,limit&&i==up,lead&&!i);
		}
	}
	if(!limit&&!lead) f[pos][len] = res;
	return res;
}
LL dp(LL n)
{
	if(!n) return 1;
	int cnt = 0;
	while(n) a[++cnt] = n%10,n/=10;
	return dfs(cnt,0,1,1);
}
int main()
{
	int T,kcase = 0;
	scanf("%d",&T);
	memset(f,-1,sizeof f);
	while(T--)
	{
		scanf("%lld%lld",&l,&r);
		if(l>r) swap(l,r); 
		printf("Case %d: %lld\n",++kcase,dp(r)-dp(l-1));
	}
	return 0;
}