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【HDOJ】4162 Shape Number

2015-12-22 11:10 295 查看
循环串的最小表示法。

/*  */
#include <iostream>
#include <string>
#include <map>
#include <queue>
#include <set>
#include <stack>
#include <vector>
#include <deque>
#include <algorithm>
#include <cstdio>
#include <cmath>
#include <ctime>
#include <cstring>
#include <climits>
#include <cctype>
#include <cassert>
#include <functional>
#include <iterator>
#include <iomanip>
using namespace std;
//#pragma comment(linker,"/STACK:102400000,1024000")

#define sti                set<int>
#define stpii            set<pair<int, int> >
#define mpii            map<int,int>
#define vi                vector<int>
#define pii                pair<int,int>
#define vpii            vector<pair<int,int> >
#define rep(i, a, n)     for (int i=a;i<n;++i)
#define per(i, a, n)     for (int i=n-1;i>=a;--i)
#define clr                clear
#define pb                 push_back
#define mp                 make_pair
#define fir                first
#define sec                second
#define all(x)             (x).begin(),(x).end()
#define SZ(x)             ((int)(x).size())
#define lson            l, mid, rt<<1
#define rson            mid+1, r, rt<<1|1

const int maxl = 3e5+5;
char M[8][8];
char s[maxl];
char d[maxl];
int len;

void init() {
rep(i, 0, 8) {
rep(j, 0, 8) {
M[i][j] = (j-i+8)%8 + '0';
}
}
}

int minStr() {
int i, j, k;
int tmp;

i = 0;
j = 1;
k = 0;
while (i<len && j<len && k<len) {
tmp = d[(i+k)%len] - d[(j+k)%len];
if (tmp == 0) {
++k;
} else {
if (tmp < 0)
j += k+1;
else
i += k+1;
if (i == j)
++j;
k = 0;
}
}

return i<j ? i:j;
}

int main() {
ios::sync_with_stdio(false);
#ifndef ONLINE_JUDGE
freopen("data.in", "r", stdin);
freopen("data.out", "w", stdout);
#endif

int k;

init();
while (scanf("%s", s) != EOF) {
len = strlen(s);
s[len] = s[0];
rep(i, 0, len) {
d[i] = M[s[i]-'0'][s[i+1]-'0'];
}
#ifndef ONLINE_JUDGE
d[len] = '\0';
puts(d);
#endif
k = minStr();
rep(i, 0, len) {
putchar(d[(k+i)%len]);
}
putchar('\n');
}

#ifndef ONLINE_JUDGE
printf("time = %d.\n", (int)clock());
#endif

return 0;
}
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