If you are new to learning programming and do not know where to start, please try Problem A “Welcome to AtCoder” from practice contest. There you can find a sample code for each language.
Also, if you are not familiar with problems in programming contests, we recommend you to try some problems in “AtCoder Beginners Selection” (https://atcoder.jp/contests/abs).

Solution 1

This problem can be solved by implementing a conditional branches with \(8\) cases. The opposite of N is S, that of SE is NW, ……, and so on. Enumerate these eight rules using if statements.

Sample code (Python 3)

D = input()
if D == "N":
    print("S")
elif D == "S":
    print("N")
elif D == "E":
    print("W")
elif D == "W":
    print("E")
elif D == "NE":
    print("SW")
elif D == "NW":
    print("SE")
elif D == "SE":
    print("NW")
elif D == "SW":
    print("NE")

Solution 2

One can flip each character of \(D\) to reduce casework to four. By using a for statement, we can handle the case where \(S\) has one character and two characters in the same way.

Sample code (Python3)

D = input()
ans = ""
for d in D:
    if d == "N":
        ans += "S"
    elif d == "S":
        ans += "N"
    elif d == "E":
        ans += "W"
    elif d == "W":
        ans += "E"
print(ans)

Sample code (C++)

#include <bits/stdc++.h>
using namespace std;

int main() {
    string D;
    cin >> D;
    string ans = "";
    for (char d : D) {
        if (d == 'N') {
            ans += 'S';
        } else if (d == 'S') {
            ans += 'N';
        } else if (d == 'E') {
            ans += 'W';
        } else if (d == 'W') {
            ans += 'E';
        }
    }
    cout << ans << endl;
}