hdu2844 Coins（多重背包）

本题要求的是1~m之间可以由所给价值组合而成的数的个数，所以dp代表的是容量，也就是正常的多重背包。所以按照模板来就可以了~

#include <stdio.h>
#include <algorithm>
#include <iostream>
#include <cmath>
#include <string.h>
using namespace std;

const int N = 500005;

int dp
, bag
, val
;
int sum;

void ZeroOnePack(int cost, int weight)
{
for(int i = sum; i >= cost; i --)
dp[i] = max(dp[i], dp[i - cost] + weight);
}

void CompletePack(int cost, int weight)
{
for(int i = cost; i <= sum; i ++)
dp[i] = max(dp[i], dp[i - cost] + weight);
}

void MulPack(int cost, int weight, int countt)
{
if(cost * countt >= sum) //Èç¹û±³°üÄÜÂú
{
CompletePack(cost, weight);
return;
}
int i = 1;
while(i < countt)
{
ZeroOnePack(i * cost, i * weight);
countt -= i;
i <<= 1;
}
ZeroOnePack(countt * cost, countt * weight);
}

int main()
{
// freopen("in.txt", "r", stdin);
int n;
while(~scanf("%d%d", &n, &sum) && n && sum)
{
memset(dp, 0, sizeof(dp));
for(int i = 0; i < n; i ++)
scanf("%d", &val[i]);
for(int i = 0; i < n; i ++)
scanf("%d", &bag[i]);
for(int i = 0; i < n; i ++)
{
MulPack(val[i], val[i], bag[i]);
}
int sum0 = 0;
for(int i = 1;i <= sum; i ++)
{
if(dp[i] == i)
{
sum0 ++;
}
}
printf("%d\n", sum0);
}
return 0;
}