Great! it worked, it is given that there would be atmost 10^4 steps that's
why i used (int). Thanks for making time to debug it. I appreciate it.
On Saturday, April 25, 2020 at 1:16:00 AM UTC+5:30, porker2008 wrote:
>
> You need to use long long to handle intermediate result, otherwise it will
> overflow.
> Otherwise, your solution looks pretty good!
>
> *#include <iostream>*
> *#include <vector>*
> *#include <string>*
> *using namespace std;*
>
> *long long MOD = 1000000000LL;*
>
> *pair<pair<long long, long long>, int> xydis(string s) {*
> * long long x = 0;*
> * long long y = 0;*
> * int i = 0;*
> * for (i = 0; s[i] != ')' && i < s.size(); i++) {*
> * if (s[i] == 'N') {*
> * y--;*
> * }*
> * if (s[i] == 'S') {*
> * y++;*
> * }*
> * if (s[i] == 'E') {*
> * x++;*
> * }*
> * if (s[i] == 'W') {*
> * x--;*
> * }*
> * if (isdigit(s[i])) {*
> * int n = s[i] - '0';*
> * pair<pair<long long, long long>, int> future = xydis(s.substr(i + 2,
> s.length() - i - 1));*
> * x += n * future.first.first;*
> * y += n * future.first.second;*
> * x = (x + MOD) % MOD;*
> * y = (y + MOD) % MOD;*
> * i += future.second + 2;*
> * }*
> * }*
> * x = (x + MOD) % MOD;*
> * y = (y + MOD) % MOD;*
> * return make_pair(make_pair(x, y), i);*
> *}*
> *int main() {*
> * int te;*
> * cin >> te;*
> * for (int i = 0; i < te; i++) {*
> * string s;*
> * cin >> s;*
> * cout << "Case #" << i + 1 << ": ";*
> * pair<int, int> result = xydis(s).first;*
> * long long x = result.first + 1;*
> * long long y = result.second + 1;*
> * cout << x << " " << y << endl;*
> * }*
> *}*
>
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