POJ1742 Coins

POJ1742 Coins

Description

People in Silverland use coins.They have coins of value A1,A2,A3...An Silverland dollar.One day Tony opened his money-box and found there were some coins.He decided to buy a very nice watch in a nearby shop. He wanted to pay the exact price(without change) and he known the price would not more than m.But he didn't know the exact price of the watch. You are to write a program which reads n,m,A1,A2,A3...An and C1,C2,C3...Cn corresponding to the number of Tony's coins of value A1,A2,A3...An then calculate how many prices(form 1 to m) Tony can pay use these coins.

Input

The input contains several test cases. The first line of each test case contains two integers n(1<=n<=100),m(m<=100000).The second line contains 2n integers, denoting A1,A2,A3...An,C1,C2,C3...Cn (1<=Ai<=100000,1<=Ci<=1000). The last test case is followed by two zeros.

Output

For each test case output the answer on a single line.

Sample Input

1
2
3
4
5
3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0

Sample Output

1
2
8
4

Answer

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
#include <cstdio>
#include <algorithm>
using namespace std;

int a[150], c[150], d[100010];;

int main() {
int n, m;
while (true) {
scanf("%d %d", &n, &m);
if (!n & !m) {
break;
}
for (int i = 0; i < n; ++i)
scanf("%d", &a[i]);
for (int i = 0; i < n; ++i)
scanf("%d", &c[i]);

fill(d, d+100010, -1);
d[0] = 0;
for (int i = 0; i < n; ++i) {
for (int j = 0; j <= m; ++j) {
if (d[j] > -1) {
d[j] = 0;
} else if (j >= a[i] && d[j - a[i]] > -1 && d[j - a[i]] < c[i]) {
d[j] = d[ j - a[i]] + 1;
}
}
}
int ans = 0;
for (int i = 1; i <= m; ++i) {
if (d[i] > -1) {
++ans;
}
}
printf("%d\n", ans);
}
return 0;
}

Analysis

  1. n 个数 a_1,a_2,...a_n,每个数有 c_1,c_2,...,c_n 个,求组和成为不大于 m 的整数个数。
  2. DP,dp[i][j] 代表前 i 个数组成数字 j,数字 i 的最少使用个数,初始值为 -1,代表前 i 个数不能组成数字 j,若 d[i-1][j] > -1,d[i][j] = 0,否则 d[i][j] = d[i][j-a[i]]+1。
  3. dp 数组只与前项有关,可以滚动压缩 dp。
# POJ

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