In a strange shop there are n types of coins of value A1, A2 ... An. You have to find the number of ways you can make K using the coins. You can use any coin at most K times.
For example, suppose there are three coins 1, 2, 5. Then if K = 5 the possible ways are:
11111
1112
122
5
So, 5 can be made in 4 ways.
Input
Input starts with an integer T (≤ 100), denoting the number of test cases.
Each case starts with a line containing two integers n (1 ≤ n ≤ 100) and K (1 ≤ K ≤ 10000). The next line contains n integers, denoting A1, A2 ... An (1 ≤ Ai ≤ 500). All Ai will be distinct.
Output
For each case, print the case number and the number of ways K can be made. Result can be large, so, print the result modulo 100000007.
Sample Input
2
3 5
1 2 5
4 20
1 2 3 4
Sample Output
Case 1: 4
Case 2: 108
題意:你有n種不同面值的硬幣,要用這些硬幣來組合成K,每種硬幣不得使用超過k次(怎麼可能超過k次)
解析:先來看第一種方法:
假設我們用a[i]來表示第i種硬幣的面額,用dp[i][j]來表示,在選用前i種硬幣的情況下,有多少種組成j的方案。
那麼可以有遞推式:dp[i][j]=dp[i-1][j-a[i]]+dp[i-1][j-2*a[i]].......dp[i-1][j-k*a[i]];
那麼我們可以嘗試着去簡化一下這個式子,雖然爲了做出這道題,沒有必要去簡化,但是爲了把我的代碼貼出來,還是需要簡化一下的。假如我們使用dp[i]來表示用當前的硬幣來組成i有多少種方案,那麼我們知道,dp[i]=dp[i-a[j]](j=1~n)但是,這樣寫遞推式,會導致112和121被視爲兩種情況,那麼我們應該如何解決這樣的問題呢,答案是,我們需要將硬幣一個個地引入,然後每引入一個新的硬幣,就去刷新一遍dp數組,下面的話看一下代碼應該就可以理解了。
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<iostream>
using namespace std;
int n,k;
int a[10000];
int dp[200000];
int main(){
int T;
cin>>T;
int cnt=0;
while(T--){
memset(a,0,sizeof(a));
memset(dp,0,sizeof(dp));
cin>>n>>k;
for(int i=1;i<=n;i++){
scanf("%d",&a[i]);
}
dp[0]=1;
sort(a+1,a+1+n);
for(int cass=1;cass<=n;cass++){
int curr=a[cass];
if(a[cass]!=a[cass-1]){
for(int i=curr;i<=k;i++){
if(curr<=i){
dp[i]+=dp[i-curr];
dp[i]%=100000007;
}
}
}
}
printf("Case %d: %d\n",++cnt,dp[k]);
}
return 0;
}