题目大意是给出一些面值的硬币和数量。问在1-m中能凑出多少种钱。
设 dp[i+1][j] 为前i种凑成j元第i种最多剩下多少。
1. dp[i+1][j] = mi ( dp[i][j]>=0) 前i-1种已经能凑成j了
2.dp[i+1][j] = -1 （ j<a[i] || dp[i+1][j-a[i]] <=0 )
3. dp[i+1][j] = dp[i+1][j-a[i]] -1
题目：
Appoint
description:
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
代码：
1 #include <iostream> 2 #include <cstring> 3 #include <cstdio> 4 using namespace std; 5 6 int n,m; 7 int A[200]; 8 int C[200]; 9 int dp[100000+10]; 10 int main() 11 { 12 while(scanf("%d%d",&n,&m)!=EOF) 13 { 14 if(n==0)break; 15 for(int i=0;i<=m;i++) 16 dp[i] = -1; 17 18 for(int i=0;i<n;i++) 19 { 20 scanf("%d",&A[i]); 21 } 22 for(int i=0;i<n;i++) 23 { 24 scanf("%d",&C[i]); 25 } 26 //dp 27 dp[0]=0; 28 for(int i=0;i<n;i++) 29 { 30 for(int j=0;j<=m;j++) 31 { 32 if( dp[j] >=0) 33 dp[j] = C[i]; 34 else if( j<A[i] || dp[j-A[i]]<=0) 35 dp[j] = -1; 36 else 37 dp[j] = dp[j-A[i]]-1; 38 } 39 } 40 int cnt =0; 41 for(int i=1;i<=m;i++) 42 { 43 if( dp[i]>=0)cnt++; 44 } 45 printf("%d\n",cnt); 46 } 47 return 0; 48 }