第一眼生成函数。四个等比数列形式的多项式相乘，可以化成四个分式。其中分母部分是固定的，可以多项式求逆预处理出来。而分子部分由于项数很少，询问时2^4算一下贡献就好了。这个思路比较直观。只是常数巨大，以及需要敲一发类似任意模数ntt的东西来避免爆精度。成功以这种做法拿下luogu倒数rank1，至于bzoj不指望能过了。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> #include<iomanip> using namespace std; int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 550000 #define T 100000 #define P1 998244353 #define P2 1004535809 int r[N],c1,c2,c3,c4,tot,d1,d2,d3,d4,s,t; int a[N],b[N],c[N],e[2][N]; long long f[N]; void inc(int &x,int P){x++;if (x>=P) x-=P;} void dec(int &x,int P){x--;if (x<0) x+=P;} int ksm(int a,int k,int P) { if (a==0) return 0; if (k==0) return 1; int tmp=ksm(a,k>>1,P); if (k&1) return 1ll*tmp*tmp%P*a%P; else return 1ll*tmp*tmp%P; } long long ksc(long long a,long long b,long long P) { long long t=a*b-(long long)((long double)a*b/P+0.5)*P; return t<0?t+P:t; } void DFT(int n,int *a,int p,int P) { for (int i=0;i<n;i++) if (i<r[i]) swap(a[i],a[r[i]]); for (register int i=2;i<=n;i<<=1) { int wn=ksm(p,(P-1)/i,P); for (register int j=0;j<n;j+=i) { int w=1; for (register int k=j;k<j+(i>>1);k++,w=1ll*w*wn%P) { int x=a[k],y=1ll*w*a[k+(i>>1)]%P; a[k]=(x+y)%P,a[k+(i>>1)]=(x-y+P)%P; } } } } void mul(int n,int *a,int *b,int P,int inv3) { DFT(n,a,3,P),DFT(n,b,3,P); for (int i=0;i<n;i++) a[i]=1ll*a[i]*(P+2-1ll*a[i]*b[i]%P)%P; DFT(n,a,inv3,P); int inv=ksm(n,P-2,P); for (int i=0;i<n;i++) a[i]=1ll*a[i]*inv%P; } void solve(int P,int inv3,int op) { memset(a,0,sizeof(a)); memset(b,0,sizeof(b)); memset(c,0,sizeof(c)); if (c1+c2+c3+c4<=T) inc(a[c1+c2+c3+c4],P); if (c1+c2+c3<=T) dec(a[c1+c2+c3],P); if (c1+c2+c4<=T) dec(a[c1+c2+c4],P); if (c4+c2+c3<=T) dec(a[c4+c2+c3],P); if (c1+c4+c3<=T) dec(a[c1+c4+c3],P); if (c1+c2<=T) inc(a[c1+c2],P); if (c1+c3<=T) inc(a[c1+c3],P); if (c1+c4<=T) inc(a[c1+c4],P); if (c2+c3<=T) inc(a[c2+c3],P); if (c4+c2<=T) inc(a[c4+c2],P); if (c3+c4<=T) inc(a[c3+c4],P); dec(a[c1],P);dec(a[c2],P);dec(a[c3],P);dec(a[c4],P); inc(a[0],P); t=1;b[0]=1; while (t<=T) { t<<=1; for (int i=0;i<t;i++) c[i]=a[i]; for (int i=0;i<(t<<1);i++) r[i]=(r[i>>1]>>1)|(i&1)*t; mul(t<<1,b,c,P,inv3); for (int i=t;i<(t<<1);i++) b[i]=0; } memcpy(e[op],b,sizeof(e[op])); } void crt() { long long P=1ll*P1*P2,inv1=ksm(P2%P1,P1-2,P1),inv2=ksm(P1%P2,P2-2,P2); for (int i=0;i<=T;i++) f[i]=(ksc(1ll*e[0][i]*P2%P,inv1,P)+ksc(1ll*e[1][i]*P1%P,inv2,P))%P; } int main() { #ifndef ONLINE_JUDGE freopen("bzoj1042.in","r",stdin); freopen("bzoj1042.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif c1=read(),c2=read(),c3=read(),c4=read(),tot=read(); solve(P1,332748118,0); solve(P2,334845270,1); crt(); while (tot--) { d1=read(),d2=read(),d3=read(),d4=read(),s=read(); d1=min(1ll*s+1,1ll*(d1+1)*c1); d2=min(1ll*s+1,1ll*(d2+1)*c2); d3=min(1ll*s+1,1ll*(d3+1)*c3); d4=min(1ll*s+1,1ll*(d4+1)*c4); long long ans=f[s]; if (d1+d2+d3+d4<=s) ans+=f[s-(d1+d2+d3+d4)]; if (d1+d2+d3<=s) ans-=f[s-(d1+d2+d3)]; if (d1+d2+d4<=s) ans-=f[s-(d1+d2+d4)]; if (d4+d2+d3<=s) ans-=f[s-(d4+d2+d3)]; if (d1+d4+d3<=s) ans-=f[s-(d1+d4+d3)]; if (d1+d2<=s) ans+=f[s-(d1+d2)]; if (d1+d3<=s) ans+=f[s-(d1+d3)]; if (d1+d4<=s) ans+=f[s-(d1+d4)]; if (d2+d3<=s) ans+=f[s-(d2+d3)]; if (d4+d2<=s) ans+=f[s-(d4+d2)]; if (d3+d4<=s) ans+=f[s-(d3+d4)]; if (d1<=s) ans-=f[s-d1]; if (d2<=s) ans-=f[s-d2]; if (d3<=s) ans-=f[s-d3]; if (d4<=s) ans-=f[s-d4]; printf(LL,ans); } return 0; }
还有一种更优秀的做法。考虑如果硬币没有个数限制的话，就是一个完全背包。添加限制可以想到容斥。我们枚举有哪几种硬币超过了个数限制，就可以容斥斥斥容容容斥把多重背包转化成完全背包了。
#include<iostream> #include<cstdio> #include<cmath> #include<cstdlib> #include<cstring> #include<algorithm> #include<iomanip> using namespace std; int read() { int x=0,f=1;char c=getchar(); while (c<'0'||c>'9') {if (c=='-') f=-1;c=getchar();} while (c>='0'&&c<='9') x=(x<<1)+(x<<3)+(c^48),c=getchar(); return x*f; } #define N 100010 #define ll long long int c[4],t,d[4],s; ll f[N],ans; int calc(int k,int x){if (k<x) return 0;else return f[k-x];} void dfs(int k,int sum,ll tot) { if (tot>s) return; if (k==4) {ans+=((sum&1)?-1:1)*f[s-tot];return;} dfs(k+1,sum+1,tot+1ll*(d[k]+1)*c[k]); dfs(k+1,sum,tot); } int main() { #ifndef ONLINE_JUDGE freopen("bzoj1042.in","r",stdin); freopen("bzoj1042.out","w",stdout); const char LL[]="%I64d\n"; #else const char LL[]="%lld\n"; #endif for (int i=0;i<4;i++) c[i]=read(); t=read(); f[0]=1; for (int i=0;i<4;i++) for (int j=c[i];j<=N-10;j++) f[j]+=f[j-c[i]]; while (t--) { for (int i=0;i<4;i++) d[i]=read(); s=read(); ans=0; dfs(0,0,0); printf(LL,ans); } return 0; }
仔细考虑一下会发现两个做法本质上其实是一样的。分子部分所乘的多项式就是一个容斥的过程，而求逆所得的结果就是完全背包。