Sicily 4874. POGODAK

2012年每週一賽第二場第二題，高效模擬問題。說白了就是一個骰子轉來轉去的問題，每次翻轉產生的變化可以自由模擬，這不是問題。問題是如果是直接模擬的話，時間複雜度是O（RC），而R和C分別都可以高達100000，它們的積就過於大了。而事實上，在每一行中翻轉時，每翻轉4次就會回到原位，那麼實際上需要模擬的就只是R*C%4次而已，這樣就可以把問題化簡了。

Run Time: 0.01sec

Run Memory: 304KB

Code Length: 1264Bytes

Submit Time: 2012-03-03 20:10:23

// Problem#: 4874
// Submission#: 1231193
// The source code is licensed under Creative Commons Attribution-NonCommercial-ShareAlike 3.0 Unported License
// URI: http://creativecommons.org/licenses/by-nc-sa/3.0/
// All Copyright reserved by Informatic Lab of Sun Yat-sen University
#include <cstdio>
#include <string>
using namespace std;

struct Cube {
    string number;
    Cube() { number = "136425"; }
    int show( int i ) { return number[ i ] - '0'; }
    void right() {
        char temp = number[ 0 ];
        number[ 0 ] = number[ 3 ];
        number[ 3 ] = number[ 2 ];
        number[ 2 ] = number[ 1 ];
        number[ 1 ] = temp;
    }
    void down() {
        char temp = number[ 0 ];
        number[ 0 ] = number[ 5 ];
        number[ 5 ] = number[ 2 ];
        number[ 2 ] = number[ 4 ];
        number[ 4 ] = temp;
    }
    void left() {
        char temp = number[ 0 ];
        number[ 0 ] = number[ 1 ];
        number[ 1 ] = number[ 2 ];
        number[ 2 ] = number[ 3 ];
        number[ 3 ] = temp;
    };
};

int main()
{
    Cube cube;
    int R, C;
    int i, j;
    long long sum = 0;

    scanf( "%d%d", &R, &C );
    for ( i = 1; i <= R; i++ ) {
        sum += C / 4 * ( cube.show( 0 ) + cube.show( 1 ) + cube.show( 2 ) + cube.show( 3 ) );
        for ( j = 1; j < C % 4; j++ ) {
            sum += cube.show( 0 );
            i % 2 == 1 ? cube.right(): cube.left();
            }
        if ( j == C % 4 )
            sum += cube.show( 0 );
        cube.down();
    }
    printf( "%lld\n", sum );

    return 0;

}