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IDProblemSubmitterResultTimeMemoryLanguageFile sizeSubmit timeJudge time
#138239#6128. Flippy Sequencesihan_88AC ✓61ms7492kbC++14456b2023-08-11 10:47:212023-08-11 10:47:25

Judging History

你现在查看的是最新测评结果

  • [2023-08-11 10:47:25]
  • 评测
  • 测评结果：AC
  • 用时：61ms
  • 内存：7492kb
查看
  • [2023-08-11 10:47:21]
  • 提交

answer

#include<cstdio>
typedef long long ll;
const int N=1000010;
char a[N],b[N];
int f[N];
int main()
{
	int T,n;
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d%s%s",&n,a+1,b+1);
		for(int i=1;i<=n;++i) f[i]=a[i]^b[i];
		int cnt=0;
		for(int i=1;i<=n;++i) cnt+=(f[i]^f[i-1]);
		cnt+=f[n];
		if(cnt==0) printf("%lld\n",((ll)n*(n+1))>>1);
		else if(cnt==2) printf("%d\n",(n-1)<<1);
		else if(cnt==4) puts("6");
		else puts("0");
	}
	return 0;
}
Source code, Language: C++14

Details

Tip: Click on the bar to expand more detailed information

Test #1:

score: 100
Accepted
time: 1ms
memory: 3644kb

input:

3
1
1
0
2
00
11
5
01010
00111

output:

0
2
6

result:

ok 3 number(s): "0 2 6"

Test #2:

score: 0
Accepted
time: 61ms
memory: 7492kb

input:

126648
1
0
0
1
1
0
2
01
01
2
01
11
2
10
11
2
11
00
3
011
011
3
010
110
3
011
001
3
111
001
3
001
000
3
101
000
3
011
000
3
111
000
4
1111
1111
4
1110
0110
4
0010
0110
4
1011
0111
4
1001
1011
4
0100
1110
4
0000
0110
4
0111
1001
4
1001
1000
4
1011
0010
4
0001
0100
4
1000
0101
4
0100
0111
4
1101
0110
4...

output:

1
0
3
2
2
2
6
4
4
4
4
6
4
4
10
6
6
6
6
6
6
6
6
6
6
6
6
6
6
6
15
8
8
8
8
6
8
8
8
6
6
6
8
6
8
8
8
6
6
6
6
0
6
6
8
6
6
6
8
6
8
8
21
10
10
10
10
6
10
10
10
6
6
6
10
6
10
10
10
6
6
6
6
0
6
6
10
6
6
6
10
6
10
10
10
6
6
6
6
0
6
6
6
0
0
0
6
0
6
6
10
6
6
6
6
0
6
6
10
6
6
6
10
6
10
10
28
12
12
12
12
6
12
12
1...

result:

ok 126648 numbers
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