Buy the souvenirs (01背包,动归)
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And you have only 7 RMB, this time you can select any combination with 3 kinds of souvenirs at most, so the selections of 3 kinds of souvenirs are ABC (6), ABD (7). But if you have 8 RMB, the selections with the most kinds of souvenirs are ABC (6), ABD (7), ACD (8), and if you have 10 RMB, there is only one selection with the most kinds of souvenirs to you: ABCD (10).
2 4 7 1 2 3 4 4 0 1 2 3 4
You have 2 selection(s) to buy with 3 kind(s) of souvenirs. Sorry, you can‘t buy anything.
All the numbers and results are in the range of 32-signed integer, and 0<=m<=500, 0<n<=30, t<=500, and the prices are all positive integers. There is a blank line between two cases.
大意:给定纪念品数目n,价格a[1~n],求m元能购买的最多纪念品数目,以及m元能购买该数目纪念品的组合数。
#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
int f[33][505],a[33];
int main()
{
int t,n,m,k,s;
scanf("%d",&t);
while(t--)
{
scanf("%d%d",&n,&m);
for(int i=1;i<=n;i++)
scanf("%d",&a[i]);
memset(f,0,sizeof(f));
sort(a+1,a+n+1);
k=n,s=0;
for(int i=1;i<=n;i++)
{
s+=a[i];
if(s>m)
{
k=i-1;
break;
}
}
for(int i=0;i<=m;i++)
f[0][i]=1;
if(k)
{
for(int i=1;i<=n;i++)
for(int j=m;j>=a[i];j--)
for(int l=1;l<=k;l++)
f[l][j]+=f[l-1][j-a[i]];
printf("You have %d selection(s) to buy with %d kind(s) of souvenirs.\n",f[k][m],k);
}
else
printf("Sorry, you can‘t buy anything.\n");
}
return 0;
}
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