You are given a string ss consisting of nn lowercase Latin letters. Let’s define a substring as a contiguous subsegment of a string. For example, “acab” is a substring of “abacaba” (it starts in position 33 and ends in position 66), but “aa” or “d” aren’t substrings of this string. So the substring of the string ss from position ll to position rr is s[l;r]=slsl+1…srs[l;r]=slsl+1…sr. You have to choose exactly one of the substrings of the given string and reverse it (i. e. make s[l;r]=srsr−1…sls[l;r]=srsr−1…sl) to obtain a string that is less lexicographically. Note that it is not necessary to obtain the minimum possible string. If it is impossible to reverse some substring of the given string to obtain a string that is less, print “NO”. Otherwise print “YES” and any suitable substring. String xx is lexicographically less than string yy, if either xx is a prefix of yy (and x≠yx≠y), or there exists such ii (1≤i≤min(|x|,|y|)1≤i≤min(|x|,|y|)), that xi<yixi<yi, and for any jj (1≤j<i1≤j<i) xj=yjxj=yj. Here |a||a| denotes the length of the string aa. The lexicographic comparison of strings is implemented by operator < in modern programming languages. Input The first line of the input contains one integer nn (2≤n≤3⋅1052≤n≤3⋅105) — the length of ss. The second line of the input contains the string ss of length nn consisting only of lowercase Latin letters. Output If it is impossible to reverse some substring of the given string to obtain a string which is lexicographically less, print “NO”. Otherwise print “YES” and two indices ll and rr (1≤l<r≤n1≤l<r≤n) denoting the substring you have to reverse. If there are multiple answers, you can print any. Examples Input 7 abacaba Output YES 2 5 Input 6 aabcfg Output NO Note In the first testcase the resulting string is “aacabba”.
一开始没读懂题，蒟蒻。 给你一个字符串，求出它是否可以倒序使字符串的字典序变小，可以输出YES和子字符串的范围，不可以输出NO。
#include<stdio.h> #include<string.h> char a[300100];//防止数组越界 注意范围。 int main() { long long n; while(~scanf("%d",&n)) { int x,y; int i; int flag=0; scanf("%s",a); for(i=0;i<strlen(a)-1;i++) if(a[i]>a[i+1]) { x=i+1; y=i+2; flag=1; break; } if(flag) {printf("YES\n"); printf("%d %d\n",x,y); } else printf("NO\n"); } return 0; }