codeforces 405 B Domino Effect (模拟)

B. Domino Effect

time limit per test

1 second

memory limit per test

256 megabytes

input

standard input

output

standard output

Little Chris knows there's no fun in playing dominoes, he thinks it's too random and doesn't require skill. Instead, he decided to play withthe dominoes and make a "domino show".

Chris arranges n dominoes in a line, placing each piece vertically upright. In the beginning, he simultaneously pushes some of the dominoes either to the left or to the right. However, somewhere between every two dominoes pushed in the same direction there is at least one domino pushed in the opposite direction.

After each second, each domino that is falling to the left pushes the adjacent domino on the left. Similarly, the dominoes falling to the right push their adjacent dominoes standing on the right. When a vertical domino has dominoes falling on it from both sides, it stays still due to the balance of the forces. The figure shows one possible example of the process.

Given the initial directions Chris has pushed the dominoes, find the number of the dominoes left standing vertically at the end of the process!

Input

The first line contains a single integer n (1 ≤ n ≤ 3000), the number of the dominoes in the line. The next line contains a character strings of length n. The i-th character of the string si is equal to

• "L", if the i-th domino has been pushed to the left;
• "R", if the i-th domino has been pushed to the right;
• ".", if the i-th domino has not been pushed.

It is guaranteed that if si = sj = "L" and i < j, then there exists such k that i < k < j and sk = "R"; if si = sj = "R" and i < j, then there exists such k that i < k < j and sk = "L".

Output

Output a single integer, the number of the dominoes that remain vertical at the end of the process.

Examples

input

14.L.R...LR..L..

output

4

input

5R....

output

0

input

1.

output

1

Note

The first example case is shown on the figure. The four pieces that remain standing vertically are highlighted with orange.

In the second example case, all pieces fall down since the first piece topples all the other pieces.

In the last example case, a single piece has not been pushed in either direction.

123

诺米骨牌模拟,  从左到右扫描,一遍,  两种状态,如果遇到L  则计数清零, 遇到R 统计.的个数

 同理, 转到 从R 开始 遇到 R 清零 遇到L  计算L_R 直接的.的个数 奇数 ans+1   最后  在统计 左右边 无R的情况 

#include <iostream>#include <stdio.h>#include <algorithm>#include <cmath>#include <cstring>#include <string>#include <queue>#include <stack>#include <stdlib.h>#include <list>#include <map>#include <set>#include <bitset>#include <vector>#define mem(a,b) memset(a,b,sizeof(a))#define findx(x) lower_bound(b+1,b+1+bn,x)-b#define FIN      freopen("input.txt","r",stdin)#define FOUT     freopen("output.txt","w",stdout)#define S1(n)    scanf("%d",&n)#define SL1(n) scanf("%I64d",&n)#define SL2(n,m) scanf("%I64d%I64d",&n,&m)#define S2(n,m)  scanf("%d%d",&n,&m)#define SS(n) scanf("%s",str)#define Pr(n)    printf("%d\n",n)using namespace std;typedef long long ll;const double PI=acos(-1);const int INF=0x3f3f3f3f;const double esp=1e-6;const int maxn=1e6+5;const int MOD=1e9+7;const int mod=1e9+7;int dir[5][2]={0,1,0,-1,1,0,-1,0};ll inv[maxn*2],fac[maxn];ll gcd(ll a,ll b){ return b?gcd(b,a%b):a;}ll exgcd(ll a,ll b,ll &x,ll &y){if(!b){x=1;y=0;return a;}ll ans=exgcd(b,a%b,x,y);ll temp=x;x=y;y=temp-a/b*y;return ans;}ll lcm(ll a,ll b){ return b/gcd(a,b)*a;}ll qpow(ll x,ll n){ll res=1;for(;n;n>>=1){if(n&1)res=(res*x)%MOD;x=(x*x)%MOD;}return res;}void INV(){inv[1] = 1;for(int i = 2; i < maxn; i++) inv[i] = (MOD - MOD / i) * inv[MOD % i] % MOD;}void ex_gcd(ll a,ll b,ll &d,ll &x,ll &y){if(!b){ x=1; y=0; d=a; }else{ ex_gcd(b,a%b,d,y,x); y-=x*(a/b);}}void Fac(){fac[0]=1;for(int i=1;i<=maxn;i++)fac[i]=(fac[i-1]*i)%MOD;}ll inv_exgcd(ll a,ll n){ll d,x,y;ex_gcd(a,n,d,x,y);return d==1?(x+n)%n:-1;}ll inv1(ll b){return b==1?1:(MOD-MOD/b)*inv1(MOD%b)%MOD;}ll inv2(ll b){return qpow(b,MOD-2);}ll cal(ll m,ll n){if(m<n)return 0;return (fac[m]*inv[fac[n]]%MOD)%MOD*inv[fac[m-n]]%MOD;}ll cals(ll m,ll n){if(m<n)return 0;return (fac[m]*inv1(fac[n])%MOD)%MOD*inv1(fac[m-n])%MOD;}char str[maxn];int a[maxn];int main(){int n;ll ans;while(~S1(n)){mem(a,0);mem(str,0);SS(str);int len=strlen(str);for(int i=0;i<len;i++){if(str[i]=='.')a[i]=0;else if(str[i]=='L')a[i]=1;else a[i]=2;}ans=0;int cont=0;int flag=0;for(int i=0;i<len;i++){if(!flag){if(a[i]==0)cont++;if(a[i]==1)cont=0;if(a[i]==2){ans+=cont;cont=0;flag=1;}}else{if(a[i]==0)cont++;if(a[i]==2)cont=0;if(a[i]==1){if(cont&1)ans++;cont=0;flag=0;}}}if(!flag)ans+=cont;Pr(ans);}}