bzoj 4052: [Cerc2013]Magical GCD
bzoj4488的双倍经验！！
1 #include <bits/stdc++.h> 2 #define LL long long 3 using namespace std; 4 inline LL ra() 5 { 6 LL x=0; char ch=getchar(); 7 while (ch<'0' || ch>'9') ch=getchar(); 8 while (ch>='0' && ch<='9') {x=x*10+ch-'0';ch=getchar();} 9 return x; 10 } 11 12 const int maxn=100005; 13 14 struct node{ 15 LL num; int x; 16 bool operator < (const node &y) const {return num<y.num||num==y.num && x<y.x;} 17 }f[2][maxn]; 18 LL a[maxn]; 19 int n; 20 21 LL gcd(LL x, LL y) {return y==0?x:gcd(y,x%y);} 22 23 int main() 24 { 25 int T=ra(); 26 while (T--) 27 { 28 n=ra(); 29 for (int i=1; i<=n; i++) a[i]=ra(); 30 int s1=0,s2=0; 31 int p=0,q=1; 32 LL ans=0; node d; 33 for (int i=1; i<=n; i++) 34 { 35 for (int j=1; j<=s1; j++) f[p][j].num=gcd(a[i],f[p][j].num); 36 d.num=a[i]; d.x=i; s1++; f[p][s1]=d; 37 sort(f[p]+1,f[p]+s1+1); 38 s2=0; 39 for (int j=1; j<=s1; j++) 40 if (f[p][j].num!=f[p][j-1].num) 41 f[q][++s2]=f[p][j]; 42 for (int j=1; j<=s2; j++) ans=max(ans,f[q][j].num*(LL)(i-f[q][j].x+1)); 43 p^=1; q^=1; s1=s2; 44 } 45 printf("%lld\n",ans); 46 } 47 return 0; 48 }
