C - Coin Change (II) Time Limit:1000MS Memory Limit:32768KB 64bit IO Format:%lld & %llu Submit Status Practice LightOJ 1232
Description
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 makeK 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) andK (1 ≤ K ≤ 10000). The next line contains n integers, denotingA1, A2 ... An (1 ≤ Ai ≤ 500). AllAi 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 modulo100000007.
Sample Input
2
3 5
1 2 5
4 20
1 2 3 4
Sample Output
Case 1: 4
Case 2: 108
题意:n种硬币,要得到k个硬币,每种硬币最多用k次,求方案数
思路:完全背包统计
#include<iostream>
#include<cstdio>
#include<cstring>
using namespace std;
int dp[10005];
int main()
{
int t,n,k,a[105],m=1;
cin>>t;
while(t--)
{
scanf("%d%d",&n,&k);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
//清零
memset(dp,0,sizeof(dp));
dp[0]=1;
//控制格式
if(m++)
printf("Case %d: ",m-1);
for(int i=1;i<=n;i++)//枚举物品的种类
{
for(int j=a[i];j<=k;j++)//枚举背包数量
{
dp[j]=(dp[j]+dp[j-a[i]])%100000007;
}
}
printf("%d\n",dp[k]);
}
return 0;
}