LightOJ 1205 - Palindromic Numbers （数位dp）

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http://www.lightoj.com/volume_showproblem.php?problem=1205

求[i,j]区间内回文数的个数。

为了使得处理到第pos位时，前面的状态是确定的，设置一个辅助数组num[ ]表示该回文数前mid位的数，根据前mid位数去确定后面的数。

这样题目就变得简单了，只需再确定了回文数的长度，然后进行记忆化。

dp[i][j]表示处理到第i位回文数的长度为j的个数。

#include <stdio.h>#include <iostream>#include <map>#include <set>#include <list>#include <stack>#include <vector>#include <math.h>#include <string.h>#include <queue>#include <string>#include <stdlib.h>#include <algorithm>//#define LL __int64#define LL long long#define eps 1e-12#define PI acos(-1.0)using namespace std;const int INF = 0x3f3f3f3f;const int maxn = 4010;int dig[20];LL dp[20][20];int num[20];LL dfs(int pos, int len, int first, int up){if(pos == 0)return 1;if(!up && !first && dp[pos][len] != -1)return dp[pos][len];int n = up ? dig[pos] : 9;LL res = 0;for(int i = 0; i <= n; i++){if(first){num[len] = i;if(i == 0)res += dfs(pos-1,len-1,1,up&&i==n);else res += dfs(pos-1,len,0,up&&i==n);}else{int mid = (len+1)>>1;if(len&1){if(pos >= mid){num[pos] = i;res += dfs(pos-1,len,first&&i==0,up&&i==n);}else{if(num[2*mid-pos] == i)res += dfs(pos-1,len,first&&i==0,up&&i==n);}}else{if(pos >= (mid+1)){num[pos] = i;res += dfs(pos-1,len,first&&i==0,up&&i==n);}else{if(num[len+1-pos] == i)res += dfs(pos-1,len,first&&i==0,up&&i==n);}}}}if(!up && !first)dp[pos][len] = res;return res;}LL cal(LL num){int len = 0;while(num){dig[++len] = num%10;num/=10;}return dfs(len,len,1,1);}int main(){int test;LL a,b;scanf("%d",&test);memset(dp,-1,sizeof(dp));for(int item = 1; item <= test; item++){scanf("%lld %lld",&a,&b);if(a > b)swap(a,b);printf("Case %d: %lld\n",item,cal(b) - cal(a-1) );}return 0;}

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