Mezo Playing Zoma
\[
Time Limit: 1 s\quad Memory Limit: 256 MB
\]
可以到达的最左是 \(-L个数\)，最右是 \(R个数\)，所以答案就是相减一下。 view
/***************************************************************
> File Name : a.cpp
> Author : Jiaaaaaaaqi
> Created Time : 2020/1/10 22:04:03
***************************************************************/
#include <bits/stdc++.h>
#define fi first
#define se second
#define pb push_back
#define pii pair<int, int>
#define dbg(x) cout << #x << " = " << (x) << endl
#define mes(a, b) memset(a, b, sizeof a)
using namespace std;
typedef long long int ll;
typedef unsigned long long int ull;
const int maxn = 1e5 + 10;
const ll mod = 1e9 + 7;
const ll INF = 1e18 + 100;
const int inf = 0x3f3f3f3f;
const double pi = acos(-1.0);
const double eps = 1e-8;
int n, m;
int cas, tol, T;
char s[maxn];
int main() {
// freopen("in", "r", stdin);
scanf("%d", &n);
scanf("%s", s+1);
int x = 0, y = 0;
for(int i=1; i<=n; i++) {
x -= s[i]=='L';
y += s[i]=='R';
}
printf("%d\n", y-x+1);
return 0;
}