H. Zebras and Ocelots -ICPC North Central NA Contest 2017

https://nanti.jisuanke.com/t/43375

简单的二进制变换

　　可以理解为当前二进制数 自增1 到2^（n-1）需要的操作数

　　注意输出格式

代码

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<bitset>
#include<cassert>
#include<cctype>
#include<cmath>
#include<cstdlib>
#include<ctime>
#include<deque>
#include<iomanip>
#include<list>
#include<map>
#include<queue>
#include<set>
#include<stack>
#include<vector>
#include <vector>
#include <iterator>
#include <utility>
#include <sstream>
#include <limits>
#include <numeric>
#include <functional>
using namespace std;
#define gc getchar()
#define mem(a) memset(a,0,sizeof(a))
//#define sort(a,n,int) sort(a,a+n,less<int>())

#define ios ios::sync_with_stdio(false);cin.tie(0);cout.tie(0);

typedef long long ll;
typedef unsigned long long ull;
typedef long double ld;
typedef pair<int,int> pii;
typedef char ch;
typedef double db;

const double PI=acos(-1.0);
const double eps=1e-6;
const ll mod=1e9+7;
const int inf=0x3f3f3f3f;
const int maxn=1e5+10;
const int maxm=100+10;

bool compare(int a, int b)
{
	return a < b;//升序
}

int main(){
	int t = 0;
	char c;
	cin >> t;
	unsigned long long num = 1;
	for(int i = 0;i<t-1;i++)
	{
		num*=2;
	}
	unsigned long long sum = 0;
	while(t--)
	{
		cin >>c;
		if(c == 'O')
		{
			sum+=num;
		}
		num /=2;
	}
	cout<<sum;
    return 0;
}