容斥原理+背包dp
每次做背包肯定会tle
那么我们可以重复利用以前的条件,先做完全背包,预处理,然后减去不合法状态
代码
//By AcerMo
#include<cmath>
#include<cstdio>
#include<cstring>
#include<iostream>
#include<algorithm>
#define lli long long int
using namespace std;
const int M=100009;
lli f[M]={1},ans;
int c[4],d[4],tot;
inline int read()
{
int x=0;char ch=getchar();
while (ch>'9'||ch<'0') ch=getchar();
while (ch>='0'&&ch<='9') x=x*10+ch-'0',ch=getchar();
return x;
}
signed main()
{
for (int i=0;i<4;i++) c[i]=read();
tot=read();
for (int i=0;i<4;i++)
for (int k=c[i];k<M;k++)
f[k]+=f[k-c[i]];
for (int i=1;i<=tot;i++)
{
for (int k=0;k<4;k++) d[k]=read();
s=read();ans=f[s];
for (int k=1;k<=15;k++)
{
int j=0,e=0,now=s;
for (int tmp=k;tmp;tmp>>=1,j++)
if(tmp&1) e^=1,now-=(d[j]+1)*c[j];
if (now>=0) e?ans-=f[now]:ans+=f[now];
}
cout<<ans<<endl;
}
return 0;
}