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QOJ
ID | Problem | Submitter | Result | Time | Memory | Language | File size | Submit time | Judge time |
---|---|---|---|---|---|---|---|---|---|
#691225 | #6128. Flippy Sequence | OIer_kzc# | AC ✓ | 46ms | 5892kb | C++17 | 828b | 2024-10-31 10:29:12 | 2024-10-31 10:29:12 |
Judging History
answer
#include <stdio.h>
#include <string.h>
#include <vector>
#include <algorithm>
#define eb emplace_back
using namespace std;
typedef long long LL;
constexpr int N = 1000005;
int n;
char a[N], b[N];
struct Pair {
int x, y;
Pair() {}
Pair(int _x, int _y) : x(_x), y(_y) {}
};
vector<Pair> seq;
int main() {
int task;
for (scanf("%d", &task); task--; ) {
scanf("%d%s%s", &n, a + 1, b + 1);
seq.clear();
for (int i = 1, j; (j = i) <= n; i = j + 1) {
while (j <= n && a[j] != b[j]) {
++j;
}
if (a[i] != b[i]) {
seq.eb(i, j - 1);
}
}
if (seq.size() >= 3) {
puts("0");
} else if (seq.size() == 2) {
puts("6");
} else if (seq.size() == 1) {
printf("%d\n", 2 * n - 2);
} else {
printf("%lld\n", n * (n + 1ll) / 2);
}
}
return 0;
}
Details
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Test #1:
score: 100
Accepted
time: 0ms
memory: 3888kb
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: 46ms
memory: 5892kb
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