ZOJ - 1729(最小表示法。 <<然后输出字典序最小
2017-07-15 11:54
218 查看
传送门
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:
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; }
相关文章推荐
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ - 1729(最小表示法。 <<然后输出字典序最小
- ZOJ 1729 Hidden Password(最小表示法模板题)
- Hidden Password ZOJ - 1729(最小表示法)模板题
- Problem Description 输入n(n<100)个数,找出其中最小的数,将它与最前面的数交换后输出这些数。 Input 输入数据有多组,每组占一行,每行的开始是一个整数n,表示这个测试实例的数值的个数,跟着就是n个整数。n=0表示输入的结束,不做处理。 Output 对于每组
- Problem Description 青年歌手大奖赛中,评委会给参赛选手打分。选手得分规则为去掉一个最高分和一个最低分,然后计算平均得分,请编程输出某选手的得分。 Input 输入数据有多组,每组占一行,每行的第一个数是n(2<n<=100),表示评委的人数,然后是n个评委的打分。 O
- ZOJ 1729 & ZOJ 2006(最小表示法模板题)
- ZOJ 1729(Hidden Password-最小表示法)
- ZOJ 3204 Connect them (最小生成树,输出字典序最小的解)
- ACM457现在给出了一个只包含大小写字母的字符串,不含空格和换行,要求把其中的大写换成小写,小写换成大写,然后输出互换后的字符串。输入 第一行只有一个整数m(m<=10),表示测试数据组数。