Coins
| Time Limit: 3000MS | Memory Limit: 30000K | |
| Total Submissions: 32309 | Accepted: 10986 |
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.
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
题意：给出3种硬币的面额和数量，能拼成不大于m的多少种；
多重背包可解，因为只要求行或不行就可以了，所以就两种状态在01和完全背包的时候没必要求可行解，只要确定行或不行就ok了，所以直接与dp[j - a[i]] 或运算，
注意的是，位运算真的好快，把dp设成int，用关系运算||,是超时的，改成位运算的|直接3000ms卡过；
改成bool型直接2204ms;
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 using namespace std; 6 const int MAX = 100000 + 10; 7 bool dp[MAX]; 8 int a[100 + 10],c[100 + 10]; 9 int n,m; 10 void ZeroOnePage(int cost) 11 { 12 for(int i = m; i >= cost; i--) 13 { 14 dp[i] |= dp[i - cost]; 15 } 16 } 17 void CompletePage(int cost, int mount) 18 { 19 for(int i = cost; i <= m; i++) 20 dp[i] |= dp[i - cost]; 21 } 22 void MultiplePage(int cost, int mount) 23 { 24 if(cost * mount >= m) 25 { 26 CompletePage(cost, mount); 27 return ; 28 } 29 int k = 1; 30 while(k < mount) 31 { 32 ZeroOnePage(k * cost); 33 mount -= k; 34 k <<= 1; 35 } 36 if(mount > 0) 37 ZeroOnePage(mount * cost); 38 return ; 39 } 40 int main() 41 { 42 while(scanf("%d%d", &n, &m) != EOF) 43 { 44 if(n == 0 && m == 0) 45 break; 46 for(int i = 1; i <= n; i++) 47 scanf("%d", &a[i]); 48 for(int i = 1; i <= n; i++) 49 scanf("%d", &c[i]); 50 memset(dp, 0, sizeof(dp)); 51 dp[0] = 1; 52 for(int i = 1; i <= n; i++) 53 if(c[i]) 54 MultiplePage(a[i], c[i]); 55 int sum = 0; 56 for(int i = 1; i <= m; i++) 57 if(dp[i]) 58 sum++; 59 printf("%d ",sum); 60 } 61 62 return 0; 63 }
#include <iostream> #include <cstring> #include <algorithm> #include <cstdio> using namespace std; const int MAX = 100000 + 10; bool dp[MAX]; int a[100 + 10],c[100 + 10]; int n,m; void ZeroOnePage(int cost) { for(int i = m; i >= cost; i--) { dp[i] |= dp[i - cost]; } } void CompletePage(int cost, int mount) { for(int i = cost; i <= m; i++) dp[i] |= dp[i - cost]; } void MultiplePage(int cost, int mount) { if(cost * mount >= m) { CompletePage(cost, mount); return ; } int k = 1; while(k < mount) { ZeroOnePage(k * cost); mount -= k; k <<= 1; } //这里是还剩下的mount if(mount > 0) ZeroOnePage(mount * cost); return ; } int main() { while(scanf("%d%d", &n, &m) != EOF) { if(n == 0 && m == 0) break; for(int i = 1; i <= n; i++) scanf("%d", &a[i]); for(int i = 1; i <= n; i++) scanf("%d", &c[i]); memset(dp, 0, sizeof(dp)); dp[0] = 1; for(int i = 1; i <= n; i++) if(c[i]) MultiplePage(a[i], c[i]); int sum = 0; for(int i = 1; i <= m; i++) if(dp[i]) sum++; printf("%d ",sum); } return 0; } 多重背包好理解
这种解法看不懂
1 #include <iostream> 2 #include <cstring> 3 #include <algorithm> 4 #include <cstdio> 5 using namespace std; 6 const int MAX = 100000 + 10; 7 int dp[MAX],used[MAX],a[100 + 10],c[100 + 10]; 8 int n,m; 9 int main() 10 { 11 while(scanf("%d%d", &n, &m) != EOF) 12 { 13 if(n == 0 && m == 0) 14 break; 15 for(int i = 1; i <= n; i++) 16 scanf("%d", &a[i]); 17 for(int i = 1; i <= n; i++) 18 scanf("%d", &c[i]); 19 memset(dp, 0, sizeof(dp)); 20 dp[0] = 1; 21 int sum = 0; 22 for(int i = 1; i <= n; i++) 23 { 24 memset(used, 0, sizeof(used)); 25 for(int j = a[i]; j <= m; j++) 26 { 27 if(dp[j] == 0 && dp[j - a[i]] && used[j - a[i]] < c[i]) 28 { 29 sum++; 30 dp[j] = 1; 31 used[j] = used[j - a[i]] + 1; 32 } 33 } 34 } 35 printf("%d ",sum); 36 } 37 38 return 0; 39 }