题意：有一个n(n<=1000)位密码锁，每位都是0~9，可以循环旋转。每次可以让1~3个相邻数字同时往上或者往下转一格。输入初始状态和终止状态(长度不超过1000),问最少要转几次。
分析：
1、从左往右依次使各个数字与终止状态相同。
2、dp[cur][x1][x2][x3]表示当前研究数字为第cur位，x1为a[cur]，x2为a[cur + 1]，x3为a[cur + 2]，在当前状态下，使所有数字变成终止状态的最小旋转次数。
3、研究第cur位时，第cur+1位和第cur+2位也可以随之一起转，枚举当前所有的旋转情况，并分别讨论向上和向下旋转两种情况。
#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
const double eps = 1e-8;
inline int dcmp(double a, double b){
if(fabs(a - b) < eps) return 0;
return a > b ? 1 : -1;
}
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const int MAXN = 1000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
char aa[MAXN], bb[MAXN];
int a[MAXN], b[MAXN];
int dp[MAXN][15][15][15];
int len;
int getUpStep(int st, int et){//向上翻转的步数，向上转数字变大
return (et - st + 10) % 10;
}
int getDownStep(int st, int et){
return (st - et + 10) % 10;
}
int upNowpos(int st, int length){
return (st + length) % 10;
}
int downNowpos(int st, int length){//从st向下翻转length步变成的数字
return (st - length + 10) % 10;
}
int dfs(int cur, int x1, int x2, int x3){
if(dp[cur][x1][x2][x3] != -1) return dp[cur][x1][x2][x3];
if(cur == len) return 0;
int ans = INT_INF;
int upstep = getUpStep(x1, b[cur]), downstep = getDownStep(x1, b[cur]);
for(int i = 0; i <= upstep; ++i){//枚举第cur+1位可以跟着第cur位一起向上旋转的步数
for(int j = 0; j <= i; ++j){//枚举第cur+2位可以跟着第cur位和第cur+1位一起向上旋转的步数
ans = Min(ans, dfs(cur + 1, upNowpos(x2, i), upNowpos(x3, j), a[cur + 3]) + upstep);
}
}
for(int i = 0; i <= downstep; ++i){//向下转
for(int j = 0; j <= i; ++j){
ans = Min(ans, dfs(cur + 1, downNowpos(x2, i), downNowpos(x3, j), a[cur + 3]) + downstep);
}
}
return dp[cur][x1][x2][x3] = ans;
}
int main(){
while(scanf("%s%s", aa, bb) == 2){
memset(dp, -1, sizeof dp);
len = strlen(aa);
for(int i = 0; i < len; ++i) a[i] = aa[i] - '0';
for(int i = 0; i < len; ++i) b[i] = bb[i] - '0';
a[len] = a[len + 1] = a[len + 2] = 0;
b[len] = b[len + 1] = b[len + 2] = 0;
printf("%d\n", dfs(0, a[0], a[1], a[2]));
}
return 0;
}