AccurateLee双指针+贪心+字符串

B. AccurateLee

time limit per test

2 seconds

memory limit per test

256 megabytes

input

standard input

output

standard output

Lee was cleaning his house for the party when he found a messy string under the carpets. Now he'd like to make it clean accurately and in a stylish way...

The string ss he found is a binary string of length nn (i. e. string consists only of 0-s and 1-s).

In one move he can choose two consecutive characters sisi and si+1si+1, and if sisi is 1 and si+1si+1 is 0, he can erase exactly one of them (he can choose which one to erase but he can't erase both characters simultaneously). The string shrinks after erasing.

Lee can make an arbitrary number of moves (possibly zero) and he'd like to make the string ss as clean as possible. He thinks for two different strings xx and yy, the shorter string is cleaner, and if they are the same length, then the lexicographically smaller string is cleaner.

Now you should answer tt test cases: for the ii-th test case, print the cleanest possible string that Lee can get by doing some number of moves.

Small reminder: if we have two strings xx and yy of the same length then xx is lexicographically smaller than yy if there is a position ii such that x1=y1x1=y1, x2=y2x2=y2,..., xi−1=yi−1xi−1=yi−1 and xi<yixi<yi.

Input

The first line contains the integer tt (1≤t≤1041≤t≤104) — the number of test cases.

Next 2t2t lines contain test cases — one per two lines.

The first line of each test case contains the integer nn (1≤n≤1051≤n≤105) — the length of the string ss.

The second line contains the binary string ss. The string ss is a string of length nn which consists only of zeroes and ones.

It's guaranteed that sum of nn over test cases doesn't exceed 105105.

Output

Print tt answers — one per test case.

The answer to the ii-th test case is the cleanest string Lee can get after doing some number of moves (possibly zero).

Example

input

Copy

5
10
0001111111
4
0101
8
11001101
10
1110000000
1
1

output

Copy

0001111111
001
01
0
1

Note

In the first test case, Lee can't perform any moves.

In the second test case, Lee should erase s2s2.

In the third test case, Lee can make moves, for example, in the following order: 11001101 →→ 1100101 →→ 110101 →→ 10101 →→ 1101 →→ 101 →→ 01.

觉得比赛的代码巨lj：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const ll mod = 1e9+7;
const double eps = 1e-8;
const ll mx = 1e6+10; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
bool isprime(int n) { if (n <= 1)return 0; for (int i = 2; i * i <= n; i++)if (n % i == 0)return 0; return 1; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(ll i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(ll i=(a),_=(b);i>=_;i--)
#define clr(a,b) memset(a, b, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pair
inline ll qpow(ll a, ll b) { return b ? ((b & 1) ? a * qpow(a * a % mod, b >> 1) % mod : qpow(a * a % mod, b >> 1)) % mod : 1; }
inline ll qpow(ll a, ll b, ll c) { return b ? ((b & 1) ? a * qpow(a * a % c, b >> 1) % c : qpow(a * a % c, b >> 1)) % c : 1; }
void ca(int kase, int ans) { cout << "Case #" << kase << ": " << ans << endl; }
//void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
ll t,m,n,k,l,r;
ll sum = 0;

//ll a[mx],b[mx];
ll cnt = 0;
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> t;
    while (t--) {
        string s;
        cin >> n;
        cin >> s;
        //ll i = 0, j = n - 1;
        ll a=-inf, b=inf;
        re(i,0,n){
            if (s[i] == '1') {
                a = i; break;
            }
        }
        for (ll i = n - 1; i >= 0; i--) {
            if (s[i] == '0') {
                b = i; break;
            }
        }
        if (a > b||a==-inf||b==inf)cout<< s;
        else {
            re(i, 0, a)cout<< s[i];
            cout<< 0;
            re(i, b+1, n)cout<< s[i];
        }
        cout << endl;
    }
    return 0;
}

优化一下：

#include<bits/stdc++.h>
using namespace std;
typedef long long ll;
const ll inf = 0x3f3f3f3f;
const ll mod = 1e9+7;
const double eps = 1e-8;
const ll mx = 1e6+10; //check the limits, dummy
typedef pair<int, int> pa;
const double PI = acos(-1);
ll gcd(ll a, ll b) { return b ? gcd(b, a % b) : a; }
ll lcm(ll a, ll b) { return a * b / gcd(a, b); }
bool isprime(int n) { if (n <= 1)return 0; for (int i = 2; i * i <= n; i++)if (n % i == 0)return 0; return 1; }
#define swa(a,b) a^=b^=a^=b
#define re(i,a,b) for(ll i=(a),_=(b);i<_;i++)
#define rb(i,a,b) for(ll i=(a),_=(b);i>=_;i--)
#define clr(a,b) memset(a, b, sizeof(a))
#define lowbit(x) ((x)&(x-1))
#define mkp make_pair
inline ll qpow(ll a, ll b) { return b ? ((b & 1) ? a * qpow(a * a % mod, b >> 1) % mod : qpow(a * a % mod, b >> 1)) % mod : 1; }
inline ll qpow(ll a, ll b, ll c) { return b ? ((b & 1) ? a * qpow(a * a % c, b >> 1) % c : qpow(a * a % c, b >> 1)) % c : 1; }
void ca(int kase, int ans) { cout << "Case #" << kase << ": " << ans << endl; }
//void sc(int& x) { scanf("%d", &x); }void sc(int64_t& x) { scanf("%lld", &x); }void sc(double& x) { scanf("%lf", &x); }void sc(char& x) { scanf(" %c", &x); }void sc(char* x) { scanf("%s", x); }
ll t,m,n,k,l,r;
ll sum = 0;
//ll a[mx],b[mx];
ll cnt = 0;
int main()
{
    ios::sync_with_stdio(false); cin.tie(0); cout.tie(0);
    cin >> t;
    while (t--) {
        cin >> n;
        string s;
        cin >> s;
        int i = 0, j = n;
        while (i < n && s[i] == '0')++i;
        while (j > 0 && s[j - 1] == '1')--j;
        if (i == j)  cout<< s << "
";
        else cout << s.substr(0, i) + '0' + s.substr(j) << "
";
    }
    return 0;
}