Coins
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
3 10
1 2 4 2 1 1
2 5
1 4 2 1
0 0
Sample Output
8
4
Source
题目类型：多重背包
算法分析：dp[i]表示容量为i的背包是否会被填满，则按照每种物品的数量进行对应的操作即可，不过总时间复杂度是O(NVlogV)的，复杂度比较高
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/************************************************* Author :supermaker Created Time :2016/4/15 21:07:41 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 = 666; int c[maxn], m[maxn], N, M; bool dp[100000+maxn]; void zero (int c) { for (int i = M; i >= c; i--) dp[i] |= dp[i-c]; } void comp (int c) { for (int i = c; i <= M; i++) dp[i] |= dp[i-c]; } void SS (int c, int m) { if (c * m >= M) comp (c); else { int k = 1; while (k < m) { zero (k * c); m -= k; k *= 2; } zero (m * c); } } int main() { //CFF; //CPPFF; while (scanf ("%d%d", &N, &M) != EOF) { if (N == 0 && M == 0) break; for (int i = 1; i <= N; i++) scanf ("%d", &c[i]); for (int i = 1; i <= N; i++) scanf ("%d", &m[i]); memset (dp, 0, sizeof (dp)); dp[0] = true; for (int i = 1; i <= N; i++) SS (c[i], m[i]); int res = 0; for (int i = 1; i <= M; i++) if (dp[i]) res++; printf ("%d\n", res); } return 0; } |
算法分析：一个复杂度是O(NV)的算法是用dp[i]表示容量为i的背包是否会被填满，sum[j]表示组成容积为j的背包所需要第i种物品的数量。最后对于每个物品遍历求解一遍即可
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/************************************************* Author :supermaker Created Time :2016/4/16 20:16:26 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 + 66; int c[166], m[166], sum[maxn]; bool dp[maxn]; 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", &c[i]); for (int i = 1; i <= N; i++) scanf ("%d", &m[i]); memset (dp, false, sizeof (dp)); dp[0] = true; int res = 0; for (int i = 1; i <= N; i++) { memset (sum, 0, sizeof (sum)); for (int j = c[i]; j <= M; j++) if (!dp[j] && dp[j-c[i]] && sum[j-c[i]] < m[i]) { dp[j] = true; sum[j] = sum[j-c[i]] + 1; res++; } } printf ("%d\n", res); } return 0; } |
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