lightoj 1205（回文数的数量 数位dp）

经典数位dp,思路就是在数的前一半的时候随意枚举，但是要考虑前导零，后一半跟前边对称就行。

#include<bits/stdc++.h>using namespace std;typedef long long ll;typedef pair<int,int> P;#define fi first#define se second#define INF 0x3f3f3f3f#define clr(x,y) memset(x,y,sizeof x)#define PI acos(-1.0)#define ITER set<int>::iteratorconst int Mod = 1e9 + 7;const int maxn = 22 + 10;int bits[maxn];ll dp[maxn][maxn][2];ll dfs(int pos,int l,bool is,bool flag)//l为长度，is判断{    if(pos < 0)return is;    if(!flag && ~ dp[pos][l][is])return dp[pos][l][is];    int up = flag ? bits[pos] : 9;    ll ret = 0;    for(int i = 0; i <= up; i ++)    {        if(l == pos && i == 0)            ret += dfs(pos - 1,l - 1,is,flag && i == up);//前导零        else if(is && pos < (l + 1)/2)ret += dfs(pos - 1,l,is && i == bits[l - pos],flag && i == up);//后一半要跟前边对称        else ret += dfs(pos - 1,l,is,flag && i == up);//前一半随意枚举    }    if(!flag)return dp[pos][l][is] = ret;    return ret;}ll calc(ll n){    int len = 0;while(n){bits[len ++] = n % 10;n /= 10;}    return dfs(len - 1,len - 1,true,true);}int main(){    clr(dp,-1);    int Tcase;scanf("%d",&Tcase);    for(int ii = 1;ii <= Tcase; ii ++)    {        ll n,m;scanf("%lld%lld",&n,&m);if(n > m){ll t = n;n = m; m = t;}        printf("Case %d: %lld\n",ii,calc(m) - calc(n - 1));    }    return 0;}