In the first example, you have to swap the third and the fourth elements, so the string becomes "aazaa". Then you have to swap the second and the third elements, so the string becomes "azaaa". So, it is possible to reverse the string in two swaps. Since the string in the second ...
string sn,cn; int ans[maxn]; int main(){ int n; cin>>n; cin>>sn; cn = sn; reverse(cn.begin(),cn.end()); for(int i=0;i<sn.size();i++){ int t = sn[i] - 'a'; que[t].push(i); } for(int i=0;i<n;i++){ int t = cn[i] - 'a'; ans[i] = que[t]...
reverse()函数 头文件#include<algorithm> string s; reverse(s.begin(),s.end());//将s反转 这两个函数让我初识了string类。。。 string可以让字符串直接比较大小,相加,复制,非常方便 AC代码: #include<iostream> #include<string> #include<algorithm> using namespace std; int main() { int T; cin...
Reverse String was rejected because it would be adjacent to the similar-styled Backspace. Penalties were too easy for the second slot (or so we thought) so somebody suggested question marks. Sadly, that change modifies the problem from cute to artificial. Testers then provided feedback that the...
E. String Reversal time limit per test2 seconds memory limit per test256 megabytes You are given a string ss. You have to reverse it — that is, the first letter should become equal to the last letter before the reversal, the second letter should become equal to the second-to-last ...
substr(l,r-l+1); t=s; reverse(t.begin(),t.end()); string a=s+'*'+t,b=t+'*'+s; string p=a.substr(0,getnext(a)); string q=b.substr(0,getnext(b)); cout<<t1<<(p.size()>q.size()?p:q)<<t2<<'\n'; } int main(){ int _; sc(_); while(_--) solve(); ...
Codeforces Round #644 (Div. 3)A->H(H为进制+reverse) http://codeforces.com/contest/1360/problem/A 题意: 用最小正方形,容下两个相同矩形。 #include<iostream> #include<cstdio> #include<algorithm> #include<cstring> using namespace std;...
(0);cin>>n;cin>>s;s=" "+s;s+=" ";for(inti=1;i<=n;++i){v[s[i]-'0'].pb(i);}reverse(s.begin(),s.end());for(inti=1;i<=n;++i){rev[i]=v[s[i]-'0'][cnt[s[i]-'0']++];}intans=0;for(inti=1;i<=n;++i){update(rev[i],1);ans+=i-get_sum(rev[i])...
设一个子串的状态为包含字符的二进制。如子串为abacdabacd,则状态为0000000000000000111100000000000000001111。 根据分析可以得到,一个子串和另一个子串如果没有交集,则两个串可以经过一次翻转合并在一起。 例如:abcdefgaabcdefga,串abab和串fgfg,可以通过翻转cdefgcdefg变成abgfedcaabgfedca。
First, Reverse traversal [1~m-1]. s1 = 0 if sum[i] < sum[m] — s1: s1 = s1 + 2 * query_max{a_i+1~m} update(the location of the maxinum, -maxinum) Then, traversal [m+1~n]. s2 = 0 if sum[i] + s2 < sum[m]: s2 = s2 — 2 * query_min{a_m+1...