ZOJ - 1729（最小表示法。 <<然后输出字典序最小

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Hidden Password

Time Limit: 2 Seconds      Memory Limit: 65536 KB
Some time the programmers have very strange ways to hide their passwords. See for example how Billy "Hacker" Geits hide his password. Billy chooses a string S composed of small Latin
letters with length L. Then he makes all L-1 one-letter left cyclic shifts of the string and takes as a password one prefix of the lexicographically first of the obtained strings (including S). For example let consider the string alabala. The cyclic one-letter
left shifts (including the initial string) are:
alabala

labalaa

abalaal

balaala

alaalab

laalaba

aalabal

and lexicographically first of them is the string aalabal. The first letter of this string is in position 6 in the initial string (the positions in the string are counted from 0).
Write a program that for given string S finds the start position of the smallest lexicographically one-letter left cyclic shift of this string. If the smallest lexicographically left
shift appears more than once then the program have to output the smallest initial position.
Your program has to be ready to solve more than one test case. The first line of the input file will contains only the number T of the test cases. Each of the following T lines will describe
one test case - first the length L of the string (5 <= L <= 100000) and then, separated by one space, the string S itself.
The output file have to contain exactly T lines with a single number each - the initial position found by your program.

Sample Input
2

6 baabaa

7 alabala

Sample Output
1

6
最小表示法

题意：
输出每个字符串的最小字典序字串的下标！

暴力：

//china no.1
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x) memset(x,0,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);cout.tie(0);
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
//const int dx[]={-1,0,1,0,-1,-1,1,1};
//const int dy[]={0,1,0,-1,1,-1,1,-1};
const int maxn=1e3+5;
const int maxx=1e5+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}
#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)

inline int Scan()
{
int res=0,ch,flag=0;
if((ch=getchar())=='-')flag=1;
else if(ch>='0' && ch<='9')res=ch-'0';
while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';
return flag ? -res : res;
}

int main()
{
int t;
close();
W(cin>>t)
{
W(t--)
{
int len;
string s;
cin>>len;
cin>>s;
//cout<<s<<endl;
string S=s;
int ans=0;
FOR(1,len,i)
{
char op=s[0];
s.erase(s.begin());
s+=op;
// cout<<s<<endl;
if(S>s)
{
S=s;
ans=i;
}
}
cout<<ans<<endl;
}
}
}

//china no.1
#include <vector>
#include <iostream>
#include <string>
#include <map>
#include <stack>
#include <cstring>
#include <queue>
#include <list>
#include <stdio.h>
#include <set>
#include <algorithm>
#include <cstdlib>
#include <cmath>
#include <iomanip>
#include <cctype>
#include <sstream>
#include <functional>
using namespace std;

#define pi acos(-1)
#define endl '\n'
#define srand() srand(time(0));
#define me(x) memset(x,0,sizeof(x));
#define foreach(it,a) for(__typeof((a).begin()) it=(a).begin();it!=(a).end();it++)
#define close() ios::sync_with_stdio(0); cin.tie(0);
typedef long long LL;
const int INF=0x3f3f3f3f;
const LL LINF=0x3f3f3f3f3f3f3f3fLL;
//const int dx[]={-1,0,1,0,-1,-1,1,1};
//const int dy[]={0,1,0,-1,1,-1,1,-1};
const int maxn=2e5+100;
const int maxx=1e5+100;
const double EPS=1e-7;
const int MOD=10000007;
#define mod(x) ((x)%MOD);
template<class T>inline T min(T a,T b,T c) { return min(min(a,b),c);}
template<class T>inline T max(T a,T b,T c) { return max(max(a,b),c);}
template<class T>inline T min(T a,T b,T c,T d) { return min(min(a,b),min(c,d));}
template<class T>inline T max(T a,T b,T c,T d) { return max(max(a,b),max(c,d));}

#define FOR(x,n,i) for(int i=x;i<=n;i++)
#define FOr(x,n,i) for(int i=x;i<n;i++)
#define W while
#define sgn(x) ((x) < 0 ? -1 : (x) > 0)

char str[maxn], tmp[maxn];
int get_minstring(char *s)
{
int len = strlen(s);
int i = 0, j = 1, k = 0;
while(i<len && j<len && k<len)
{
int t=s[(i+k)%len]-s[(j+k)%len];
if(t==0)
k++;
else
{
if(t > 0)
i+=k+1;
else
j+=k+1;
if(i==j) j++;
k=0;
}
}
return min(i,j);
}
int main()
{
int t,len;
scanf("%d",&t);
while(t--)
{
scanf("%d",&len);
scanf("%s",str);
int ans = get_minstring(str);
printf("%d\n",ans);
}
return 0;
}