poj3181 (Dollar Dayz)

Description

Farmer John goes to Dollar days at the Cow Store and discovers a unlimited number of tools on sale. During his first visit, the tools is selling variously for $, $, and $. Farmer John has exactly $ to spend. He can buy 5 tools at $1 tool at $ and a additional 1 tool at $. Of course, there is other combinations for a total of 5 different ways FJ can spend all he money on tools. Here they is:

1 @ us$3 + 1 @ us$2  1 @ us$3 + 2 @ us$1  1 @ us$2 + 3 @ us$1  2 @ us$2 + 1 @ us$1  5 @ us$1

Write a program than would compute the number of ways FJ can spend N dollars (1 <= n <=) at the Cow Store for OLS on sale with a cost of $ $K (1 <= K <= 100).

Input

A single line with the space-separated integers:n and K.

Output

A single, with a single integer, is the number of the unique ways FJ can spend.

Sample Input

5 3

Sample Output

5

Test instructions: How many ways to make an n in an integer in 1~m.

Method 1: Complete knapsack problem, pay attention to high precision.

Method 2: I'm using dynamic programming to split integers, and it's easier to think of them. Set Dp[n][m] Indicates the number of methods of n with an integer of not more than M, discussed in the following case:

M=0, dp[n][0]=0;

M>n, Dp[n][m]=dp[n][n];

0<m<=n, the number of methods is dp[n-m][m] If the largest integer is M, and if the maximum integer is less than M, the method number is dp[n][m-1]. So the total is dp[n][m]=dp[n-m][m]+dp[n][m-1].

The space can also be further optimized by using a scrolling array minus the second dimension.

The last is to pay attention to high precision, the large number is divided into two parts, the right half belongs to a long long range, the left half divided by 1e18 after another save, see Code.

#include <iostream> #include <sstream> #include <fstream> #include <string> #include <map > #include <vector> #include <list> #include <set> #include <stack> #include <queue># Include <deque> #include <algorithm> #include <functional> #include <iomanip> #include < limits> #include <new> #include <utility> #include <iterator> #include <cstdio> #include < cstdlib> #include <cstring> #include <cctype> #include <cmath> #include <ctime>using Namespace Std;typedef Long Long ll;const LL MOD = 1e18;    LL a[1005], B[1005];int main () {int n, m;    CIN >> n >> m;    A[0] = 1; for (int j = 1, J <= m; ++j) for (int i = j; i <= N; ++i) {b[i] = B[i-j] + b[i] + (A[I-J)            + a[i])/MOD;        A[i] = (A[i-j] + a[i])% MOD;    } if (B[n]) printf ("%lld", B[n]);    printf ("%lld\n", A[n]); return 0;}

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poj3181 (Dollar Dayz)