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BZOJ 2301: [HAOI2011]Problem b

2019-11-08 02:59:00
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首先对于a≤x≤b,c≤y≤d这样的限制,可以容斥一下,转换成1≤x≤n这样 设N=⌊nk⌋,M=⌊mk⌋ ∑ni=1∑mj=1,(i,j)=k1−> ∑Ni=1∑Mj=1[(i,j)=1]−> 然后反演一下−> ∑Ni=1∑Mj=1∑d|(i,j)μ(d)−> ∑dμ(d)∑⌊Nd⌋i=1,∑⌊mk⌋j=11−> ∑dμ(d)⌊Nd⌋⌊mk⌋

O(nn√)

code:

#include<set>#include<map>#include<deque>#include<queue>#include<stack>#include<cmath>#include<ctime>#include<bitset>#include<string>#include<vector>#include<cstdio>#include<cstdlib>#include<cstring>#include<climits>#include<complex>#include<iostream>#include<algorithm>#define ll long long#define lowbit(x) x&(-x)using namespace std;void up(int &x,int y){if(x<y)x=y;}void down(int &x,int y){if(x>y)x=y;}const int maxn = 101000;int p[maxn],PRi,miu[maxn],s[maxn];bool v[maxn];void pre(){ memset(v,false,sizeof v); miu[1]=s[1]=1; pri=0; for(int i=2;i<maxn;i++) { if(!v[i]) { miu[i]=-1; p[++pri]=i; } for(int j=1;j<=pri;j++) { int k=i*p[j]; if(k>=maxn) break; v[k]=true; if(i%p[j]==0) { miu[k]=0; break; } miu[k]=-miu[i]; } s[i]=s[i-1]+miu[i]; }}int solve(int n,int m,int k){ int re=0; n/=k; m/=k; k=min(n,m); int r; for(int l=1;l<=k;l=r+1) { r=min(n/(n/l),m/(m/l)); re+=(s[r]-s[l-1])*(n/l)*(m/l); } return re;}int main(){ pre(); int t; scanf("%d",&t); while(t--) { int a,b,c,d,k; scanf("%d%d%d%d%d",&a,&b,&c,&d,&k); int a1=solve(b,d,k); int a2=solve(a-1,d,k); int a3=solve(b,c-1,k); int a4=solve(a-1,c-1,k); printf("%d/n",a1-a2-a3+a4); } return 0;}
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