POJ3181 Dollar Dayz

POJ3181 Dollar Dayz

Description

Farmer John goes to Dollar Days at The Cow Store and discovers an unlimited number of tools on sale. During his first visit, the tools are selling variously for $1, ​$2, and $3. Farmer John has exactly $5 to spend. He can buy 5 tools at $1 each or 1 tool at $3 and an additional 1 tool at $2. Of course, there are other combinations for a total of 5 different ways FJ can spend all his money on tools. Here they are:

1
2
3
4
5
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 will compute the number of ways FJ can spend N dollars (1 <= N <= 1000) at The Cow Store for tools on sale with a cost of \(1..\)K (1 <= K <= 100).

Input

A single line with two space-separated integers: N and K.

Output

A single line with a single integer that is the number of unique ways FJ can spend his money.

Sample Input

1
5 3

Sample Output

1
5

Answer

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
#include <cstdio> 

#define LIMIT_ULL 100000000000000000
unsigned long long dp[1050][2];

int main() {
int n, k;
scanf("%d %d", &n, &k);

dp[0][1] = 1;
for (int i = 1; i <= k; ++i) {
for (int j = i; j <= n; ++j) {
dp[j][0] += dp[j-i][0];
dp[j][1] += dp[j-i][1];
dp[j][0] += dp[j][1] / LIMIT_ULL;
dp[j][1] = dp[j][1] % LIMIT_ULL;
}
}
if (dp[n][0])
printf("%lld", dp[n][0]);
printf("%lld\n", dp[n][1]);
return 0;
}

Analysis

  1. 农夫约翰有N元钱,市场上有价值1……K的商品无限个,求所有的花钱方案?
  2. 完全背包 + 大数相加
# POJ

Comments

Your browser is out-of-date!

Update your browser to view this website correctly. Update my browser now

×