Codeforces Round #786 (Div. 3) B. Dictionary

time limit per test2 seconds
memory limit per test512 megabytes
inputstandard input
outputstandard output
The Berland language consists of words having exactly two letters. Moreover, the first letter of a word is different from the second letter. Any combination of two different Berland letters (which, by the way, are the same as the lowercase letters of Latin alphabet) is a correct word in Berland language.

The Berland dictionary contains all words of this language. The words are listed in a way they are usually ordered in dictionaries. Formally, word a comes earlier than word b in the dictionary if one of the following conditions hold:

the first letter of a is less than the first letter of b;
the first letters of a and b are the same, and the second letter of a is less than the second letter of b.
So, the dictionary looks like that:

Word 1: ab
Word 2: ac
…
Word 25: az
Word 26: ba
Word 27: bc
…
Word 649: zx
Word 650: zy
You are given a word s from the Berland language. Your task is to find its index in the dictionary.

Input
The first line contains one integer t (1≤t≤650) — the number of test cases.

Each test case consists of one line containing s — a string consisting of exactly two different lowercase Latin letters (i. e. a correct word of the Berland language).

Output
For each test case, print one integer — the index of the word s in the dictionary.

Example
inputCopy
7
ab
ac
az
ba
bc
zx
zy
outputCopy
1
2
25
26
27
649
650

#include<iostream>
using namespace std;
int main(void) {
	int n;
	scanf("%d", &n);
	getchar();
	char a[3];
	for (int i = 0; i < n; i++) {
		gets(a);
		if (a[1] > a[0]) {
			printf("%d\n", (a[0] - 'a') * 25 + (a[1] - 'a'));
		}
		else {
			printf("%d\n", (a[0] - 'a') * 25 + (a[1] - 'a')+1);
		}
	}
	return 0;
}