QOJ.ac

QOJ

ID题目提交者结果用时内存语言文件大小提交时间测评时间
#235440#2831. Game Theoryyiyiyi#WA 56ms3888kbC++144.2kb2023-11-02 19:26:042023-11-02 19:26:04

Judging History

你现在查看的是最新测评结果

  • [2023-11-02 19:26:04]
  • 评测
  • 测评结果:WA
  • 用时:56ms
  • 内存:3888kb
  • [2023-11-02 19:26:04]
  • 提交

answer

#include<cstdio>
#include<cstdlib>
#include<cstring>
#include<cmath>
#include<iostream>
#include<iomanip>
#include<algorithm>
#include<vector>
#include<map>
#include<queue>
#include<bitset>
#include<set> 
#define ll long long
#define lowbit(x) x&(-x)
#define mp make_pair
#define rep(i,x,n) for(int i=x;i<=n;i++)
#define per(i,n,x) for(int i=n;i>=x;i--)
#define forE(i,x) for(int i=head[x];i;i=nxt[i])
#define pii pair<int,int>
#define fi first
#define se second
using namespace std;
const int maxn=2e5+5;
const int mod=998244353;
inline int read()
{
	int x=0,f=1;char c=getchar();
	while(c<'0'||c>'9')
	{
		if(c=='-') f=-1;
		c=getchar();
	}
	while(c>='0'&&c<='9')
	{
		x=x*10+(c-'0');
		c=getchar();
	}
	return x*f;
}

char c[maxn];

struct SegTree{
	struct node{
		int l,r;
        ll num[2],sum[2],add;
	}t[maxn<<2];
	inline void spread(int p)
	{
		if(t[p].add)
		{
			t[p*2].add^=t[p].add;
			swap(t[p*2].num[0],t[p*2].num[1]);
            swap(t[p*2].sum[0],t[p*2].sum[1]);
            t[p*2+1].add^=t[p].add;
			swap(t[p*2+1].num[0],t[p*2+1].num[1]);
            swap(t[p*2+1].sum[0],t[p*2+1].sum[1]);
			t[p].add=0;
		}
	}
	inline void pushup(int p)
	{
        t[p].sum[0]=t[p*2].sum[0]+t[p*2+1].sum[0];
        t[p].sum[1]=t[p*2].sum[1]+t[p*2+1].sum[1];
        t[p].num[0]=t[p*2].num[0]+t[p*2+1].num[0];
        t[p].num[1]=t[p*2].num[1]+t[p*2+1].num[1];
    }
	inline void build(int p,int l,int r)
	{
		t[p].l=l;t[p].r=r;t[p].add=0,t[p].sum[0]=t[p].sum[1]=t[p].num[0]=t[p].num[1]=0;
		if(l==r)
		{
            int x=c[l]-'0';
            t[p].num[x]++;
            t[p].sum[x]+=l;
			return;
		}
		int mid=(t[p].l+t[p].r)/2;
		build(p*2,l,mid);
		build(p*2+1,mid+1,r);
		pushup(p);
	}
	inline void change(int p,int l,int r)
	{
		if(t[p].l>=l&&t[p].r<=r)
		{
			swap(t[p].num[0],t[p].num[1]);
            swap(t[p].sum[0],t[p].sum[1]);
			t[p].add^=1;
			return;
		}
		spread(p);
		int mid=(t[p].l+t[p].r)/2;
		if(l<=mid) change(p*2,l,r);
		if(r>mid) change(p*2+1,l,r);
		pushup(p);
	}
	inline ll asknum(int p,int l,int r,int v)
	{
        if(l>r) return 0;
		if(t[p].l>=l&&t[p].r<=r)
		{
			return t[p].num[v];
		}
		spread(p);
		int mid=(t[p].l+t[p].r)/2;
		ll res=0;
		if(l<=mid) res+=asknum(p*2,l,r,v);
		if(r>mid) res+=asknum(p*2+1,l,r,v);
		return res;
	}
	inline ll asksum(int p,int l,int r,int v)
	{
        if(l>r) return 0;
		if(t[p].l>=l&&t[p].r<=r)
		{
			return t[p].sum[v];
		}
		spread(p);
		int mid=(t[p].l+t[p].r)/2;
		ll res=0;
		if(l<=mid) res+=asksum(p*2,l,r,v);
		if(r>mid) res+=asksum(p*2+1,l,r,v);
		return res;
	}    
    inline ll get1(int p,int v)
    {
        if(t[p].l==t[p].r) return t[p].l;
        spread(p);
        int mid=(t[p].l+t[p].r)/2;
        if(t[p*2+1].num[1]>=v) return get1(p*2+1,v);
        else return get1(p*2,v-t[p*2+1].num[1]);
    }
    inline ll get0(int p,int v)
    {
        if(t[p].l==t[p].r) return t[p].l;
        spread(p);
        int mid=(t[p].l+t[p].r)/2;
        if(t[p*2].num[0]>=v) return get0(p*2,v);
        else return get0(p*2+1,v-t[p*2].num[0]);
    }
}S; 
int n,q;

signed main()
{
    while(scanf("%d %d",&n,&q) != EOF)
    {
        scanf("%s",c+1);
        S.build(1,1,n);
        rep(i,1,q)
        {
            int p1=read(),p2=read();
            S.change(1,p1,p2);
            int tot = S.asknum(1,1,n,0);
            if(tot==0) {printf("%d\n",n);continue;}
            if(tot==n) {printf("0\n");continue;}  
            int l=1,r=n,res=-1;
            while(l<=r)
            {
                int mid=(l+r)/2;
                int x=S.asknum(1,1,mid,0),y=S.asknum(1,mid+1,n,1);
                if(x<=y) res=mid,l=mid+1;
                else r=mid-1;
            }
            if(S.asknum(1,1,res,0) == 0) {printf("%lld\n",S.asknum(1,1,n,1));continue;}
            ll num=S.asknum(1,1,res,0);
            ll L=S.get0(1,num);
            ll R=S.get1(1,num);
            // printf("3: %lld\n",res);
            ll ans=S.asknum(1,1,L,1);
            // printf("4: %lld %lld %lld\n",num,L,R);
            ans+=(S.asksum(1,R,n,1)-S.asksum(1,1,L,0)+num)*2ll-num;
            printf("%lld\n",ans);
        }
    }
}

詳細信息

Test #1:

score: 100
Accepted
time: 0ms
memory: 3884kb

input:

3 2
010
1 2
2 3
5 1
00000
1 5

output:

1
3
5

result:

ok 3 lines

Test #2:

score: 0
Accepted
time: 0ms
memory: 3876kb

input:

1 1
0
1 1

output:

1

result:

ok single line: '1'

Test #3:

score: 0
Accepted
time: 26ms
memory: 3888kb

input:

2 2
01
2 2
2 2
2 2
01
1 2
1 2
1 2
1
1 1
1 1
1 2
1
1 1
1 1
2 2
00
1 2
1 2
2 2
11
1 2
1 2
2 2
01
2 2
1 1
2 2
10
2 2
1 2
2 2
01
1 2
1 2
1 2
0
1 1
1 1
2 2
01
1 2
2 2
1 2
0
1 1
1 1
1 2
1
1 1
1 1
2 2
10
1 2
1 1
1 2
0
1 1
1 1
2 2
01
1 2
1 2
2 2
10
1 2
1 1
1 2
0
1 1
1 1
1 2
0
1 1
1 1
1 2
1
1 1
1 1
2 2
10
1 ...

output:

0
3
1
3
0
1
0
1
2
0
0
2
0
1
2
0
1
3
1
0
1
2
1
0
0
1
3
2
1
0
1
3
3
2
1
0
1
0
0
1
0
1
0
2
2
1
0
1
2
1
1
0
2
0
2
3
1
0
0
1
2
0
0
1
0
1
0
1
1
0
1
2
2
0
0
2
0
1
0
1
1
0
1
0
1
0
0
1
0
1
0
1
2
0
3
0
1
0
0
1
1
0
1
0
1
2
0
2
1
0
0
3
1
2
0
1
3
2
1
0
0
1
2
0
2
0
0
1
0
1
1
0
1
0
1
0
1
0
1
2
1
0
3
1
0
3
0
1
0
1
...

result:

ok 200000 lines

Test #4:

score: -100
Wrong Answer
time: 56ms
memory: 3808kb

input:

11 20
00010111000
1 6
1 11
5 6
8 11
1 4
3 8
4 11
1 7
5 9
1 4
6 10
5 6
1 7
5 10
1 10
9 11
6 8
1 4
8 11
1 4
10 20
0101000010
3 4
2 2
4 8
4 6
6 7
3 7
3 3
1 5
1 5
6 8
2 2
2 4
2 6
5 7
1 3
2 5
6 8
7 9
5 8
3 10
4 20
1011
2 4
1 4
1 3
2 3
1 1
2 2
1 2
2 4
1 2
3 4
3 4
3 4
1 4
2 2
1 4
1 3
1 1
1 3
1 3
2 2
4 20
1...

output:

16
55
51
25
12
15
43
49
38
32
30
40
43
45
35
26
24
30
44
36
17
20
30
38
36
39
36
29
36
26
17
31
17
17
25
19
21
10
12
30
2
10
5
6
10
7
6
2
0
10
0
10
2
1
7
6
7
4
7
7
4
7
1
5
7
3
5
7
3
7
6
7
6
7
6
6
2
0
5
3
64
103
82
68
92
137
92
75
85
92
127
119
119
104
122
62
70
86
52
52
40
43
21
21
32
59
67
32
29
50...

result:

wrong answer 3rd lines differ - expected: '53', found: '51'