题目：
Description
You have a string ss of length nn 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−1n−1, but not the whole string). You need to calculate the minimum number of characters to be deleted from string ss so that it becomes good.
Input
The first line contains one integer tt (1≤t≤1001≤t≤100) – the number of test cases. Each test case is represented by two lines.
The first line of ii-th test case contains one integer nn (1≤n≤1001≤n≤100) – the length of string ss.
The second line of ii-th test case contains string ss, consisting of only characters > and <.
Output
For each test case print one line.
For ii-th test case print the minimum number of characters to be deleted from string ss so that it becomes good.
Example
Input
3
2
<>
3
><<
1
>
Output
1
0
0
题意分析：
第一次居然理解错了题目的意思，以为是字符串先用它给定的操作变成不能再变的情况，然后我再人工删除
实际上是我先人工删除，然后再用它给定的操作一次性删除完
那么需要最后能一次性全部删除完，对于'>'可以删除这个字符右边的字符（如果'>'在最左边，可以一次性把右边所有的全部删除）
对于'<'可以删除这个字符左边的字符（如果'<'在最右边，也可以将其左边所有的字符全部删除)
那么我只需要寻找最左边的'>'跟最右边的'<'，然后比较两者需要进行人工更改的字符最少的
代码：
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int t,x,y,ans,n;
char s[105];
bool flag;
int main()
{
scanf("%d",&t);
while(t--)
{
scanf("%d",&n);
scanf("%s",s);
if(s[0]=='>')printf("0\n");
else if(s[n-1]=='<')printf("0\n");
else
{
for(int i=0;i<n;i++)
{
if(s[i]=='>')//寻找最左边的>，那么需要人工删除的字符个数为x
{
x=i;
break;
}
}
for(int i=0;i<n;i++)
{
if(s[i]=='<')y=i;//寻找最右边的<，需要删除的人工字符个数为n-(y+1)
}
ans=min(x,n-y-1);//比较两种方法下寻找人工删除的字符个数最少的
printf("%d\n",ans);
}
}
return 0;
}