Timus : 1017 (Staircases)



// http://acm.timus.ru/problem.aspx?space=1&num=1017
// Runtime: 0.078s
// Tag: Dp

// @BEGIN_OF_SOURCE_CODE

#include <iostream>
#include <cstdio>
#include <algorithm>
#include <cstring>
#include <string>
#include <cctype>
#include <stack>
#include <queue>
#include <list>
#include <vector>
#include <map>
#include <sstream>
#include <cmath>
#include <bitset>
#include <utility>
#include <set>
#include <numeric>

#define INF_MAX 2147483647
#define INF_MIN -2147483647
#define pi acos(-1.0)
#define N 1000005
#define LL long long

#define For(i, a, b) for( int i = (a); i < (b); i++ )
#define Fors(i, sz) for( size_t i = 0; i < sz.size (); i++ )
#define Fore(it, x) for(typeof (x.begin()) it = x.begin(); it != x.end (); it++)
#define Set(a, s) memset(a, s, sizeof (a))
#define Read(r) freopen(r, "r", stdin)
#define Write(w) freopen(w, "w", stdout)

int dr [] = {-1, -1, 0, 1, 1, 1, 0, -1};
int dc [] = {0, 1, 1, 1, 0, -1, -1, -1};

using namespace std;

LL memo [500 + 5] [500 + 5];

void reset ()
{
    for ( int i = 0; i < 505; i++ )
        for ( int j = 0; j < 505; j++ ) memo [i] [j] = -1;
}

LL dp (int prev, int remain)
{
    if ( remain == 0 ) return 1;
    if ( remain <= prev ) return 0;

    if ( memo [prev] [remain] != -1 ) return memo [prev] [remain];

    LL ret = 0;

    for ( int i = prev + 1; i <= remain; i++ ) ret += dp (i, remain - i);

    return memo [prev] [remain] = ret;
}

int main ()
{
    int n;

    while ( scanf ("%d", &n) != EOF ) {
        reset ();
        printf ("%lld\n", dp (0, n) - 1);
    }

	return 0;
}

// @END_OF_SOURCE_CODE
Advertisements

4 thoughts on “Timus : 1017 (Staircases)

  1. @NS
    at first you need to understand dp problems and solving criteria
    u can take help from topcoder algorithms tutorial

    if you how to apply dp then here is the hints:

    line 51:
    prev = number of bricks you have set on previous column
    remain = number of remaining bricks

    line 53:
    if remain = 0, then you have got a solution

    line 54:
    if remain <= prev, then you cant get a solution

    line 56:
    if you had come to this point before then don't calculate one more time, just return the previous calculated result

    line 60:
    now u r trying to fill up the current column with every possible number of bricks
    obviously started from prev + 1 to <= remain

    line 71:
    prev = 0
    remain = n
    dp (0, n) gives the final result
    but solutions with one column not acceptable
    so we subtracted 1 from the final result

Leave a Reply

Fill in your details below or click an icon to log in:

WordPress.com Logo

You are commenting using your WordPress.com account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s