CF 1140B 思维

B. Good String
You have a string s of length n consisting of only characters > and <. You may do some operations with this string, for each operation you have to choose some character that still remains in the string. If you choose a character >, the character that comes right after it is deleted (if the character you chose was the last one, nothing happens). If you choose a character <, the character that comes right before it is deleted (if the character you chose was the first one, nothing happens).
For example, if we choose character > in string > > < >, the string will become to > > >. And if we choose character < in string > <, the string will become to <.
The string is good if there is a sequence of operations such that after performing it only one character will remain in the string. For example, the strings >, > > are good.
Before applying the operations, you may remove any number of characters from the given string (possibly none, possibly up to n−1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string s so that it becomes good.
Input
The first line contains one integer t (1≤t≤100) – the number of test cases. Each test case is represented by two lines.
The first line of i-th test case contains one integer n (1≤n≤100) – the length of string s.
The second line of i-th test case contains string s, consisting of only characters > and <.
Output
For each test case print one line.
For i-th test case print the minimum number of characters to be deleted from string s so that it becomes good.
Example
input
3
2
<>
3
><<
1
>
output
1
0
0
Note
In the first test case we can delete any character in string <>.
In the second test case we don’t need to delete any characters. The string > < < is good, because we can perform the following sequence of operations: > < < → < < → <.
代码：

#include
#define maxn 10005
using namespace std;
typedef long long ll;
inline int read(){
    	int x=0,f=0;char ch=getchar();
    	while(ch>'9'||ch<'0')f|=ch=='-',ch=getchar();
    	while(ch>='0'&&ch<='9')x=(x<<3)+(x<<1)+(ch^48),ch=getchar();
    	return f?-x:x;
}
string s;
int main(){
	int t=read();
	while(t--){
		int len=read();
		cin>>s;
		if(len==1){
			puts("0");
			continue;
		}
		int a=0,b=0,l=0,r=len-1;
		while(s[l]=='<')++a,++l;
		while(s[r]=='>')++b,--r;
		printf("%d\n",min(a,b));
	}
	return 0;
}