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18 lines
455 B
18 lines
455 B
#include <bits/stdc++.h>
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using namespace std;
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const int N = 100010;
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int ans = 1;//最少一种中序排列
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string a, B;
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int main() {
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cin >> a >> b;
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for (int i = 0; i < a.size() - 1; i++) {
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string t = a.substr(i, 2); // 截取相邻的两个字符串
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reverse(t.begin(), t.end()); // 反转
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if (b.find(t) != string::npos) ans *= 2; // 匹配成功则总数*2
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}
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cout << ans << endl;
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return 0;
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} |