2016 ACM/ICPC Dalian Online-1002 Different GCD Subarray Query
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题意:给定长度为N的数组,Q个询问,每个询问求区间[L,R]中gcd的种类数 (1≤N,Q≤100000,1≤ai≤1000000)
题解:首先对于以a[i]为右边界的gcd种类数不超过log(a[i]),
因为每次取gcd时要么不变(种类不增加),要么质因子个数减少(种类+1),由于质因子个数最多log(a[i])个,所以种类数不超过log(a[i])
我们记录下每个a[i]的不同gcd值及最右边位置。然后对询问区间按右边界从小到大排序,用树状数组T[i]记录i位置的不同gcd个数(相同的gcd只记录最右边的一个),pos[i]表示gcd=i的最右边位置。暴力枚举右边界,右边界从i->i+1时,更新树状数组,回答询问。
#include <iostream>#include <cstdio>#include <cstring>#include <cstdlib>#include <cmath>#include <algorithm>using namespace std;const int N=1e5+10;const int M=1e6+5;struct Node{ int x,y,n,Ans;}c[N];int n,Q,a[N];int f[22][N],p[22][N];int T[N],pos[M];bool cmp(Node i,Node j){return i.y<j.y;}bool cmp2(Node i,Node j){return i.n<j.n;}int gcd(int a,int b){return b==0?a:gcd(b,a%b);}void Add(int x,int n,int v){ for (;x<=n;x+=x&(-x))T[x]+=v;}int Sum(int x){ int Ans=0; for (;x;x-=x&(-x))Ans+=T[x]; return Ans;}int main(){ //freopen("1.txt","r",stdin); while (scanf("%d%d",&n,&Q)!=EOF){ for (int i=1;i<=n;i++)scanf("%d",&a[i]); for (int i=1;i<=Q;i++)scanf("%d%d",&c[i].x,&c[i].y),c[i].n=i; memset(f[0],0,sizeof f[0]); for (int i=1;i<=n;i++){ for (int j=1;j<=f[0][i-1];j++){ int g=gcd(a[i],f[j][i-1]); if (f[f[0][i]][i]!=g){ f[0][i]++;f[f[0][i]][i]=g;p[f[0][i]][i]=p[j][i-1]; }else p[f[0][i]][i]=p[j][i-1]; } if (f[f[0][i]][i]!=a[i])f[0][i]++,f[f[0][i]][i]=a[i],p[f[0][i]][i]=i; else p[f[0][i]][i]=i; } /*for (int i=1;i<=n;i++){ printf("%d %d:",i,a[i]); for (int j=1;j<=f[0][i];j++)printf("%d ",p[j][i]); puts(""); }*/ sort(c+1,c+Q+1,cmp); memset(T,0,sizeof T); memset(pos,0,sizeof pos); int cNow=1; for (int i=1;i<=n;i++){ for (int j=1;j<=f[0][i];j++){ if (pos[f[j][i]]==0){ pos[f[j][i]]=p[j][i]; Add(p[j][i],n,1); }else if (pos[f[j][i]]<p[j][i]){ Add(pos[f[j][i]],n,-1); pos[f[j][i]]=p[j][i]; Add(p[j][i],n,1); } } while (cNow<=Q && c[cNow].y==i){ c[cNow].Ans=Sum(i)-Sum(c[cNow].x-1); cNow++; } } sort(c+1,c+Q+1,cmp2); for (int i=1;i<=Q;i++){ printf("%d\n",c[i].Ans); } } return 0;}
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