這兩天刷了兩道過去的原題，看看思維還是8太行。

這道題問給出兩個字串，要求只翻轉一次，問有幾種不同的方法使得a串變成b串

我一開始沒看到只翻轉一次，還以為是個計數 + 字串dp大難題，心想當年的學長隊伍真厲害啊能上去拿1血，結果仔細看了看發現是個水題，只轉一次，那就記錄最大相等的串，然後翻過來看相等不相等，然後向外向內拓展看有沒有頭 == 尾的情況就行了，水題一道，8多說

#include <bits/stdc++.h>
using namespace std;
#define limit (20000 + 5)//防止溢位
#define INF 0x3f3f3f3f
#define inf 0x3f3f3f3f3f
#define
 
 lowbit(i) i&(-i)//一步兩步
#define EPS 1e-6
#define FASTIO  ios::sync_with_stdio(false);cin.tie(0);
#define ff(a) printf("%d\n",a );
#define pi(a,b) pair<a,b>
#define rep(i, a, b) for(ll i = a; i <= b ; ++i)
#define per(i, a, b) for(ll i = b ; i >= a ; --i)
#define MOD 998244353
#define traverse(u) for(int i = head[u]; ~i ; i = edge[i].next)
#define
 
 FOPEN freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\data.txt", "rt", stdin)
#define FOUT freopen("C:\\Users\\tiany\\CLionProjects\\acm_01\\dabiao.txt", "wt", stdout)
#define debug(x) cout<<x<<endl
typedef long long ll;
typedef unsigned long long ull;
inline ll read(){
    ll sign = 1, x = 0
 
;char s = getchar();
    while(s > '9' || s < '0' ){if(s == '-')sign = -1;s = getchar();}
    while(s >= '0' && s <= '9'){x = (x << 3) + (x << 1) + s - '0';s = getchar();}
    return x * sign;
}//快讀
void write(ll x){
    if(x < 0) putchar('-'),x = -x;
    if(x / 10) write(x / 10);
    putchar(x % 10 + '0');
}
int f[limit];
int ans = 1;
string str,str1;
void checkin(int l, int r){
    ++l,--r;
    while (l < r){
        if(str[l] == str[r])++ans, --r, ++l;
        else return;
    }
}
int n;
void checkout(int l, int r){
    --l,++r;
    while (l >= 1 && r <= n){
        if(str[l] == str[r])++ans, ++r, --l;
        else return;
    }
}
int main() {
#ifdef LOCAL
    FOPEN;
#endif

    cin>>str>>str1;
    n = str.length();
    str = ' ' + str;
    str1 = ' ' + str1;
    int l  = -1, r = -1;
    rep(i ,1,n){
        if(str[i] != str1[i]) {
            if(l == -1){
                l = i;

            }
            r = max(i, 1ll * r);
        }
    }
    string fst;
    string scd;
    rep(i ,l,r){
        fst += str[i];
        scd += str1[i];
    }
    reverse(scd.begin(), scd.end());
    if(fst != scd)return 0 * puts("0");
    checkin(l,r);
    checkout(l,r);
    write(ans);
    return 0;
}