如何将字符串的子串反转，用于Samara大学生程序设计竞赛问题E？
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本文共计547个文字，预计阅读时间需要3分钟。
给定两个长度相同的字符串s和t，判断是否可以通过对s中某个子串进行一次反转，恰好得到t。输入：第一行包含字符串s，第二行包含字符串t。
Two strings s and t of the same length are given. Determine whether it is possible to make t from s using
exactly one reverse of some its substring.
Input
The first line contains the string s, and the second — the string t. Both strings have the same length from
1 to 200000 characters and consist of lowercase Latin letters.
Output
Output «YES», if it is possible to reverse some substring of s to make s equal to t, and «NO», otherwise.
Examples
standard input standard output
abcdefg
abedcfg
YES
abcdefg
abdecfg
NO
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 500005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-7
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
a = a % mod;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
bool prime(int x) {
if (x == 0 || x == 1)return false;
for (int i = 2; i <= sqrt(x); i++) {
if (x%i == 0)return false;
}
return true;
}
int main()
{
ios::sync_with_stdio(false);
string s, t;
cin >> s >> t;
int i, j;
int len = s.length();
int pos1, pos2;
if (s == t) {
int flag = 0;
cout << "YES" << endl;
}
else {
for (i = 0; i < len; i++) {
if (s[i] != t[i]) {
pos1 = i;
break;
}
}
for (i = len - 1; i >= 0; i--) {
if (s[i] != t[i]) {
pos2 = i;
break;
}
}
int pp1 = pos1, pp2 = pos2;
int flag = 0;
while (1) {
if (s[pos1] != t[pos2]) {
flag = 1;
break;
}
pos1++;
pos2--;
if (pos1 > pp2)break;
}
if (flag)cout << "NO" << endl;
else cout << "YES" << endl;
}
}
本文共计547个文字，预计阅读时间需要3分钟。
给定两个长度相同的字符串s和t，判断是否可以通过对s中某个子串进行一次反转，恰好得到t。输入：第一行包含字符串s，第二行包含字符串t。
Two strings s and t of the same length are given. Determine whether it is possible to make t from s using
exactly one reverse of some its substring.
Input
The first line contains the string s, and the second — the string t. Both strings have the same length from
1 to 200000 characters and consist of lowercase Latin letters.
Output
Output «YES», if it is possible to reverse some substring of s to make s equal to t, and «NO», otherwise.
Examples
standard input standard output
abcdefg
abedcfg
YES
abcdefg
abdecfg
NO
#include<iostream>
#include<cstdio>
#include<algorithm>
#include<cstdlib>
#include<cstring>
#include<string>
#include<cmath>
#include<map>
#include<set>
#include<vector>
#include<queue>
#include<string>
#include<bitset>
#include<ctime>
#include<deque>
typedef long long ll;
using namespace std;
typedef unsigned long long int ull;
#define maxn 500005
#define ms(x) memset(x,0,sizeof(x))
#define Inf 0x7fffffff
#define inf 0x3f3f3f3f
const long long int mod = 1e9 + 7;
#define pi acos(-1.0)
#define pii pair<int,int>
#define eps 1e-7
#define pll pair<ll,ll>
ll quickpow(ll a, ll b) {
ll ans = 1;
a = a % mod;
while (b > 0) {
if (b % 2)ans = ans * a;
b = b / 2;
a = a * a;
}
return ans;
}
int gcd(int a, int b) {
return b == 0 ? a : gcd(b, a%b);
}
bool prime(int x) {
if (x == 0 || x == 1)return false;
for (int i = 2; i <= sqrt(x); i++) {
if (x%i == 0)return false;
}
return true;
}
int main()
{
ios::sync_with_stdio(false);
string s, t;
cin >> s >> t;
int i, j;
int len = s.length();
int pos1, pos2;
if (s == t) {
int flag = 0;
cout << "YES" << endl;
}
else {
for (i = 0; i < len; i++) {
if (s[i] != t[i]) {
pos1 = i;
break;
}
}
for (i = len - 1; i >= 0; i--) {
if (s[i] != t[i]) {
pos2 = i;
break;
}
}
int pp1 = pos1, pp2 = pos2;
int flag = 0;
while (1) {
if (s[pos1] != t[pos2]) {
flag = 1;
break;
}
pos1++;
pos2--;
if (pos1 > pp2)break;
}
if (flag)cout << "NO" << endl;
else cout << "YES" << endl;
}
}