QOJ.ac

QOJ

IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#55589#1289. A + B ProblemCrysflyAC ✓142ms6352kbC++112.7kb2022-10-14 18:45:222022-10-14 18:45:25

Judging History

你现在查看的是最新测评结果

  • [2023-08-10 23:21:45]
  • System Update: QOJ starts to keep a history of the judgings of all the submissions.
  • [2022-10-14 18:45:25]
  • 评测
  • 测评结果:AC
  • 用时:142ms
  • 内存:6352kb
  • [2022-10-14 18:45:22]
  • 提交

answer

// what is matter? never mind.
#include<bits/stdc++.h>
#define For(i,a,b) for(int i=(a);i<=(b);++i)
#define Rep(i,a,b) for(int i=(a);i>=(b);--i)
using namespace std;
inline int read()
{
    char c=getchar();int x=0;bool f=0;
    for(;!isdigit(c);c=getchar())f^=!(c^45);
    for(;isdigit(c);c=getchar())x=(x<<1)+(x<<3)+(c^48);
    if(f)x=-x;return x;
}

// modint
#define mod 998244353
struct modint{
	int x;
	modint(int o=0){x=o;}
	modint &operator = (int o){return x=o,*this;}
	modint &operator +=(modint o){return x=x+o.x>=mod?x+o.x-mod:x+o.x,*this;}
	modint &operator -=(modint o){return x=x-o.x<0?x-o.x+mod:x-o.x,*this;}
	modint &operator *=(modint o){return x=1ll*x*o.x%mod,*this;}
	modint &operator ^=(int b){
		modint a=*this,c=1;
		for(;b;b>>=1,a*=a)if(b&1)c*=a;
		return x=c.x,*this;
	}
	modint &operator /=(modint o){return *this *=o^=mod-2;}
	friend modint operator +(modint a,modint b){return a+=b;}
	friend modint operator -(modint a,modint b){return a-=b;}
	friend modint operator *(modint a,modint b){return a*=b;}
	friend modint operator /(modint a,modint b){return a/=b;}
	friend modint operator ^(modint a,int b){return a^=b;}
	friend bool operator ==(modint a,int b){return a.x==b;}
	friend bool operator !=(modint a,int b){return a.x!=b;}
	bool operator ! () {return !x;}
	modint operator - () {return x?mod-x:0;}
	bool operator <(const modint&b)const{return x<b.x;}
};
inline modint qpow(modint x,int y){return x^y;}

vector<modint> fac,ifac,iv;
inline void initC(int n)
{
	if(iv.empty())fac=ifac=iv=vector<modint>(2,1);
	int m=iv.size(); ++n;
	if(m>=n)return;
	iv.resize(n),fac.resize(n),ifac.resize(n);
	For(i,m,n-1){
		iv[i]=iv[mod%i]*(mod-mod/i);
		fac[i]=fac[i-1]*i,ifac[i]=ifac[i-1]*iv[i];
	}
}
inline modint C(int n,int m){
	if(m<0||n<m)return 0;
	return initC(n),fac[n]*ifac[m]*ifac[n-m];
}
inline modint sign(int n){return (n&1)?(mod-1):(1);}

#define fi first
#define se second
#define pb push_back
#define mkp make_pair
typedef pair<int,int>pii;
typedef vector<int>vi;

#define maxn 1500005
#define inf 0x3f3f3f3f

int n,m,a[maxn],t,s[maxn],pos[maxn];
char str[maxn];
int res[maxn];
void work()
{
	n=read(),t=0,m=read();
	For(i,0,n+m+1) res[i]=0;
	cin>>str+1;
	For(i,1,n+m)a[i]=(str[i]&1),s[i]=s[i-1]+a[i],(a[i])&&(pos[++t]=i);
	int l=n,r=m;
	For(i,1,n+m){
		if(l>r)swap(l,r);
		if(a[i]){
			if(l && s[i]+r-l+1<=s[n+m] && pos[s[i]+r-l+1]-pos[s[i]]-(r-l+1)<l) ++res[l],--l;
			else ++res[r],--r;
		}
		else{
			if(l)--l;
			else --r;
		}
	}
	n=max(n,m);
	For(i,1,n)res[i+1]+=res[i]/2,res[i]&=1;
	++n;
	while(!res[n]&&n>1)--n;
	Rep(i,n,1)cout<<res[i];puts("");
}

signed main()
{
	int T=read();
	while(T--)work();
	return 0;
}

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 0ms
memory: 3696kb

input:

3
4 3
1000101
2 2
1111
1 1
00

output:

1101
110
0

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 142ms
memory: 6352kb

input:

11110
10 8
111011010011100100
3 5
01011000
7 6
1110101010000
9 1
0110100101
1 9
0100001110
8 10
000101101011111000
9 6
011111111000111
1 9
1011101101
10 7
00100011000100000
4 9
1000101101010
8 4
100100110000
8 9
00101111011000101
8 9
11000000101011110
7 6
1111010100110
2 9
01001110101
4 5
100010100
...

output:

10011010100
11100
10101000
110100101
100001110
10000001100
1000010111
111101101
1110100000
111101010
11110000
1000011101
1001011110
10101110
101110101
11100
1111010
1000010
1011100010
10010101001
10010001
1001010
1000000010
1110
111
1111110001
10110111
1100010101
10000000
111000011
110
11111
1100101...

result:

ok 11110 lines