# Sol

$c$只有$4$种，可以容斥一波

# include <bits/stdc++.h>
# define RG register
# define IL inline
# define Fill(a, b) memset(a, b, sizeof(a))
using namespace std;
typedef long long ll;
const int _(1005);

IL int Input(){
RG int x = 0, z = 1; RG char c = getchar();
for(; c < '0' || c > '9'; c = getchar()) z = c == '-' ? -1 : 1;
for(; c >= '0' && c <= '9'; c = getchar()) x = (x << 1) + (x << 3) + (c ^ 48);
return x * z;
}

typedef int Arr[_];
int c[4], tot, mx, pw[4] = {1, 2, 4, 8}, s[_], d[4][_];
ll f[_ * 100] = {1}, ans;

int main(RG int argc, RG char *argv[]){
for(RG int i = 0; i < 4; ++i) c[i] = Input();
tot = Input();
for(RG int i = 1; i <= tot; ++i){
for(RG int j = 0; j < 4; ++j) d[j][i] = (Input() + 1) * c[j];
s[i] = Input(), mx = max(mx, s[i]);
}
for(RG int i = 0; i < 4; ++i)
for(RG int j = c[i]; j <= mx; ++j) f[j] += f[j - c[i]];
for(RG int i = 1; i <= tot; ++i){
ans = 0;
for(RG int j = 0; j < 16; ++j){
RG int op = 1, sum = 0;
for(RG int k = 0; k < 4; ++k)
if(j & pw[k]) op = -op, sum += d[k][i];
ans += (sum > s[i]) ? 0 : (1LL * op * f[s[i] - sum]);
}
printf("%lld\n", ans);
}
return 0;
}


posted @ 2018-03-27 20:42  Cyhlnj  阅读(...)  评论(...编辑  收藏