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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#96935 | #6128. Flippy Sequence | zhuozhuo | AC ✓ | 63ms | 13456kb | C++14 | 886b | 2023-04-15 19:45:26 | 2023-04-15 19:45:27 |
Judging History
answer
#include <bits/stdc++.h>
using namespace std;
#define int long long
#define N 1000010
int same[N];
char s1[N],s2[N];
signed main()
{
int t;cin>>t;
while(t--)
{
int n;scanf("%lld",&n);
scanf("%s%s",s1,s2);
for(int i=0;i<n;i++)
{
if(s1[i]==s2[i]) same[i]=1;
else same[i]=0;
}
int cnt=0,flag=0;//记录不同区间的个数,是否出现不同
//找区间数量
for(int i=0;i<n;i++)
{
if(!flag&&!same[i]) {flag=1;cnt++;}
else if(flag&&!same[i]) ;
else if(flag&&same[i]) flag=0;
}
if(cnt==0) cout<<n*(n+1)/2<<endl;
if(cnt==1)//只有一个区间
printf("%lld\n",(n-1)*2);
if(cnt==2)//有两个区间
puts("6");
if(cnt>2) {puts("0");continue;}
}
}
Details
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Test #1:
score: 100
Accepted
time: 2ms
memory: 3520kb
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: 63ms
memory: 13456kb
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