題干：

BaoBao has just found two strings??and??in his left pocket, where??indicates the?-th character in string?, and??indicates the?-th character in string?.

As BaoBao is bored, he decides to select a substring of??and reverse it. Formally speaking, he can select two integers??and??such that??and change the string to?.

In how many ways can BaoBao change??to??using the above operation exactly once? Let??be an operation which reverses the substring?, and??be an operation which reverses the substring?. These two operations are considered different, if??or?.

Input

There are multiple test cases. The first line of the input contains an integer?, indicating the number of test cases. For each test case:

The first line contains a string??(), while the second line contains another string??(). Both strings are composed of lower-cased English letters.

It's guaranteed that the sum of??of all test cases will not exceed?.

Output

For each test case output one line containing one integer, indicating the answer.

Sample Input

2 abcbcdcbd abcdcbcbd abc abc

Sample Output

3 3

Hint

For the first sample test case, BaoBao can do one of the following three operations: (2, 8), (3, 7) or (4, 6).

For the second sample test case, BaoBao can do one of the following three operations: (1, 1), (2, 2) or (3, 3).

題目大意：

給你兩個串s1和s2,可以翻轉s1串的一個區間，只能翻轉一次，,問有多少對l,r使得翻轉后的s1串等于s2串

解題報告：

? 當兩串完全相同的時候就是馬拉車，不同的時候就先兩邊往里掃到兩串第一個不相同的位置，在第一個串往外擴，看能擴幾次，答案就是多少。

注意無數細節，，，那個初始化（見注釋）還有那個while(1)里面必須是if(l>=ed)就可以break了一開始寫成了l>ed，，，WA了一上午難受難受。。。

其實不是細節多，，而是代碼姿勢不太對，，，還是太菜，，不過這么簡單的一題比賽的時候沒有開有點小虧。做網絡同步賽打了三小時比賽水了7題，就去準備晚上給學弟講課的PPT去了。。。這成績對比了下，好像7題放現場賽也只是個銅（不過罰時較少可能是銀？）。。不過這題是真不難。。

AC代碼：

#include<cstdio> #include<iostream> #include<algorithm> #include<queue> #include<map> #include<vector> #include<set> #include<string> #include<cmath> #include<cstring> #define F first #define S second #define ll long long #define pb push_back #define pm make_pair using namespace std; typedef pair<int,int> PII; const int MAX = 4e6 + 5; char s1[MAX],s2[MAX],str[MAX<<1]; int p[MAX<<1],top; int manacher() {if(top == 0) return 0 ;int maxx = -1;int c = -1,r = -1;for(int i = 1; i<top; i++) {p[i] = r>i ? min(p[2*c-i],r-i) : 1;for(;str[i-p[i]] == str[i+p[i]];p[i]++);if(i+p[i] > r) {r = i+p[i];c=i;}maxx = max(maxx,p[i]);}return maxx-1; }int main() {int t;cin>>t;while(t--) {scanf("%s%s",s1,s2);int len = strlen(s1);if(strcmp(s1,s2) != 0) {int st=-1,ed=len,l,r;//需要賦初值！！！ int flag = 1;for(int i = 0; i<len; i++) {if(s1[i] == s2[i]) st = i;else break;}for(int i = len-1; i>=0; i--) {if(s1[i] == s2[i]) ed = i;else break;}l=st+1,r=ed-1;while(1) {if(l >= ed) break;if(s1[l] == s2[r]) l++,r--;else {flag = 0;break; }}if(flag == 0) {puts("0");continue; }l=st,r=ed; while(1) {if(l==-1||r==len) break;if(s1[l] == s1[r]) l--,r++;else break; }printf("%d\n",st-l+1); continue;}top=0;str[top++] = '@';for(int i = 0; i<len; i++) {str[top++] = '#';str[top++] = s1[i];}str[top++] = '#';str[top] = '\0';manacher();ll ans = 0;for(int i = 2; i<top; i++) { ans += p[i]/2; }printf("%lld\n",ans);}return 0 ; } //@#a#a#a# //@#a#a#

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總結

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