Coins
Problem Description
Whuacmers use coins.They have coins of value A1,A2,A3...An Silverland dollar. One day Hibix opened purse 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
3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0
Sample Output
8
4
Source
2009 Multi-University Training Contest 3 - Host by WHU
题目类型：多重背包
算法分析：dp[j]表示前i种硬币恰好组成价格为j的方案数，则按照状态转移方程dp[j] += dp[j-c[i]]跑一个多重背包，最后只要在1~m中看有多少个dp[i]大于0即可
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/************************************************* Author :supermaker Created Time :2015/11/24 0:21:40 File Location :C:\Users\abcd\Desktop\TheEternalPoet **************************************************/ #pragma comment(linker, "/STACK:102400000,102400000") #include <set> #include <bitset> #include <list> #include <map> #include <stack> #include <queue> #include <deque> #include <string> #include <vector> #include <ios> #include <iostream> #include <fstream> #include <sstream> #include <iomanip> #include <algorithm> #include <utility> #include <complex> #include <numeric> #include <functional> #include <cmath> #include <ctime> #include <climits> #include <cstdarg> #include <cstdio> #include <cstdlib> #include <cstring> #include <cctype> #include <cassert> using namespace std; #define CFF freopen ("aaa.txt", "r", stdin) #define CPPFF ifstream cin ("aaa.txt") #define DB(ccc) cout << #ccc << " = " << ccc << endl #define PB push_back #define MP(A, B) make_pair(A, B) typedef long long LL; typedef unsigned long long ULL; typedef double DB; typedef pair <int, int> PII; typedef pair <int, bool> PIB; const int INF = 0x7F7F7F7F; const int MOD = 1e9 + 7; const double EPS = 1e-10; const double PI = 2 * acos (0.0); const int maxn = 1e5 + 6666; int dp[maxn], aa[maxn], cc[maxn]; void PP (int V, int c, int num) { if (c * num >= V) { for (int i = c; i <= V; i++) dp[i] += dp[i-c]; } int k = 1; while (k < num) { for (int i = V; i >= k * c; i--) dp[i] += dp[i-k*c]; num -= k; k <<= 1; } for (int i = V; i >= num * c; i--) dp[i] += dp[i-num*c]; } int main() { //CFF; //CPPFF; int n, m; while (scanf ("%d%d", &n, &m) != EOF) { if (n == 0 && m == 0) break; for (int i = 1; i <= n; i++) scanf ("%d", &aa[i]); for (int i = 1; i <= n; i++) scanf ("%d", &cc[i]); memset (dp, 0, sizeof (dp)); dp[0] = 1; for (int i = 1; i <= n; i++) PP (m, aa[i], cc[i]); int res = 0; for (int i = 1; i <= m; i++) if (dp[i]) res++; printf ("%d\n", res); } return 0; } |
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