B. JoJo‘s Incredible Adventures(双指针)

Problem - B - Codeforces

给定一个长度为n的01字符串， 每次将整个串整体往后移，最后一个移到开头，作为下一行，问存在的全1矩形的最大面积

输入

5
0
1
101
011110
101010
 

输出

0
1
2
6
1
 

如果一行中最多的连续1的个数为x， 则只有一行矩形的面积为x * 1 ，两行的面积为（x-1） * 2 ,三行的面积是(x-2) * 3 ,最终面积有一个最大值，也就是（x-i+ 1）* i,i为x除2上取整

#include <iostream>
#include <vector>
#include <bits/stdc++.h>
#include <unordered_map>
#include <unordered_set>
#include <queue>
#include <set>
#include <stack>
#include <algorithm>
#define x first
#define y second
#define pb emplace_back
#define fu(i,a,b) for(int i=a;i<=b; ++ i)
#define fd(i,a,b) for(int i=a;i>=b;	-- i)
#define endl '\n'
#define ms(x,y) memset(x,y,sizeof x)
#define ios ios::sync_with_stdio(0),cin.tie(0),cout.tie(0);
using namespace std;typedef long long LL;
typedef unsigned long long ULL;
typedef vector<vector<LL>> VVL;
typedef vector<vector<int>> VVI;
typedef vector<LL> VL;
typedef vector<bool> VB;
typedef vector<int> VI;
typedef vector<string> VS;
typedef pair<int,int> PII;
typedef vector<PII> VPII;
typedef pair<PII,int> PIII;
typedef pair<double,double> PDD;
typedef pair<double,int> PDI;
typedef pair<char,int> PCI;
typedef pair<string,int> PSI;
typedef pair<int,string> PIS;
typedef pair<int,char> PIC;
typedef pair<LL,LL> PLL;
typedef __int128 i128;
typedef unsigned long long ULL;
const int N =40 + 10 ,M = N *100 ,INF = 0x3f3f3f3f,P = 131;
const double eps = 1e-8,DNF = 1e18;
const int mod = 1e9 + 7 ,base= 20010;
const LL LNF=(LL) INF * INF;int n;inline void solve()
{string s;cin >> s;if(count(s.begin(),s.end(),'0') == 0){cout << s.size()  * s.size() << endl;return;}int n= s.size();s=' ' + s + s;// s -> 变成循环，长度* 2  int len=0;for(int i=1,j;i  < s.size() ; i++ )if(s[i] =='1')	{j = i ;while(j < s.size() && s[j] == '1') j ++ ;len = max(len,min(n,j-i));i = j-1 ;}// len 最长连续1 的个数 cout << 1ll * (len + 1 >> 1) * (len +  1 - (len + 1 >> 1 )) << endl;}signed main()
{
//  freopen("1.txt","r",stdin);
//	#define int long long 
//	init(N-1);ios
//	cout << fixed<<setprecision(2);int t=1;cin>>t;int now = 1;while(t -- ){
//      cout<<"Case "; 
//      cout<<"Case #"; 
//      cout<<"Scenario #"; 
//      cout<< now ++ <<": ";
//      cout<< now ++ <<": \n";solve();}return 0;
}