Codeforces Round #465 (Div. 2) B. Fafa and the Gates

B. Fafa and the Gates

Two neighboring kingdoms decided to build a wall between them with some gates to enable the citizens to go from one kingdom to another. Each time a citizen passes through a gate, he has to pay one silver coin.

The world can be represented by the first quadrant of a plane and the wall is built along the identity line (i.e. the line with the equation x = y). Any point below the wall belongs to the first kingdom while any point above the wall belongs to the second kingdom. There is a gate at any integer point on the line (i.e. at points (0, 0), (1, 1), (2, 2), ...). The wall and the gates do not belong to any of the kingdoms.

Fafa is at the gate at position (0, 0) and he wants to walk around in the two kingdoms. He knows the sequence S of moves he will do. This sequence is a string where each character represents a move. The two possible moves Fafa will do are 'U' (move one step up, from (x, y) to (x, y + 1)) and 'R' (move one step right, from (x, y) to (x + 1, y)).

Fafa wants to know the number of silver coins he needs to pay to walk around the two kingdoms following the sequence S. Note that if Fafa visits a gate without moving from one kingdom to another, he pays no silver coins. Also assume that he doesn't pay at the gate at point (0, 0), i. e. he is initially on the side he needs.

Input

The first line of the input contains single integer n (1 ≤ n ≤ 105) — the number of moves in the walking sequence.

The second line contains a string S of length n consisting of the characters 'U' and 'R' describing the required moves. Fafa will follow the sequence S in order from left to right.

Output

On a single line, print one integer representing the number of silver coins Fafa needs to pay at the gates to follow the sequence S.

Examples

input

Copy

1
U

output

0

input

Copy

6
RURUUR

output

1

input

Copy

7
URRRUUU

output

2

题意：在第一象限内，有一条y=x的直线，有两个操作：U向上一步，R向右一步。现在起点位于0,0，给你一系列操作，问你会有多少次跨过这条y=x的线。

思路：首先可以判断，两步之内必然不能跨线，所以直接从第三个i开始，与i-2进行匹配，中线是y=x，那么如果跨线，连续的三个坐标必然是：xy 或者：x>y   x==y   x

#include
#define ll long long
using namespace std;
struct node
{
    ll x,y;
}a[100010];
char s[100010];
ll n,ans,sx,sy;
int main()
{
    while(~scanf("%lld",&n))
    {
        scanf("%s",s+1);
        ans=sx=sy=0;
        for(ll i=1;i<=n;i++)
        {
            if(s[i]=='U')sy++;
            else sx++;
            a[i].x=sx;
            a[i].y=sy;
            if(i>=3)
            {
                if(a[i].x>a[i].y&&a[i-2].xa[i-2].y)ans++;
            }
        }
        printf("%lld\n",ans);
    }
    return 0;
}