[HAOI2008]硬币购物
Description:
\ \ 方程$a_1 x_1+a_2x_2+a_3x_3+a_4x_4=s, $\ \ \(a_1\),\(a_2\),\(a_3\),\(a_4\)给定
\ \ 对于 \(q\) 组 \(d_1\)~\(d_4\) 和 \(s\) 求满足\(x_1<=d_1\), \(x_2<=d_2\), \(x_3<=d_3\), \(x_4<=d_4\)正整数解的个数
Hint:
\ \ \(d_i<=1e5\), \(s<=1e5\), \(q<=1e3\)
solution:
\ \ 多重背包显然超时
\ \ 考虑补集转化,求所有正整数解的个数-不满足di限制的个数
\ \ 设\(S_i\)为\(x_i>d_i\)的解的个数,\(f[x]\)为\(x\)时的正整数解
\ \ 即求\(S_1∪S_2∪S_3∪S_4=∑S_i-∑S_i∩S_j+∑S_i∩S_j∩S_k-\)......
\ \ 可以\(2^4\)枚举\(s-(d_i+1)*a_i\)......的\(f[]\)和,再用\(f[s]\)减去即可
\ \ 至于\(f[]\),一遍预处理即可
#include<bits/stdc++.h>
#define long long int
using namespace std;
const int mxn=1e5+5;
int f[mxn],a[5],d[5],t,s;
signed main()
{
scanf("%d %d %d %d %d",&a[1],&a[2],&a[3],&a[4],&t);
f[0]=1;
for(int i=1;i<=4;++i)
for(int j=a[i];j<=mxn-5;++j)
f[j]+=f[j-a[i]]; //无限背包预处理
for(int i=1;i<=t;++i) {
scanf("%d %d %d %d %d",&d[1],&d[2],&d[3],&d[4],&s);
int ans=0;
for(int j=0;j<16;++j) { //枚举子集
int cnt=0,tp=s;
for(int k=0;k<4;++k)
if((j>>k)&1) ++cnt,tp-=(d[k+1]+1)*a[k+1];
if(tp<0) continue ;
if(cnt&1) ans-=f[tp]; //容斥
else ans+=f[tp];
}
printf("%lld\n",ans);
}
return 0;
}