感觉是个炒鸡高级的算法。。。由于本人太渣了。。。只做了2题QAQ
%%%lbn187 _(:зゝ∠)_
lbn上课讲了CDQ分治,莫队,还有各种奇奇怪怪的离线问题
感觉莫队和分块好像啊。。。
我之前做过一道(捂脸)分块的题目哎。。。
BZOJ-2957
(我猜泥萌肯定没想到竟然有人拿之前的程序来充数。。。
#include <iostream>
#include <cstdio>
#include <cmath>
#include <cstring>
#include <cstdlib>
#define I int
#define D double
#define NN 100005
#define NNN 330
using namespace std;
struct REC
{
int l;
D num[NNN];
void add(D xx) {num[++l]=xx;}
int finds(D xx)
{
int left=1,right=l,tmp=0;
while (left<=right)
{
int mid=(left+right)>>1;
if (num[mid]>xx) tmp=mid,right=mid-1; else left=mid+1;
}
return tmp;
}
};
REC q[NN];
int n,m,x,y,len,cnt,now,ans,lenn[NN];
D w[NN],maxx;
int main()
{
scanf("%d%d",&n,&m);
len=(I)sqrt((D)n);
while (len*len>n) len--;
cnt=n/len+(n%len==0?0:1);
for (int i=1;i<cnt;++i) lenn[i]=len;
lenn[cnt]=n-(cnt-1)*len;
for (int i=1;i<=m;++i)
{
scanf("%d%d",&x,&y);
w[x]=(D)y/(D)x;
now=(x-1)/len+1;
maxx=0; q[now].l=0;
for (int j=len*(now-1)+1;j<=len*(now-1)+lenn[now];++j)
if (w[j]>maxx)
q[now].add(w[j]),maxx=w[j];
ans=q[1].l;
maxx=q[1].num[ans];
for (int j=2;j<=cnt;++j)
if (q[j].num[q[j].l]>maxx)
{
ans+=q[j].l-q[j].finds(maxx)+1;
maxx=q[j].num[q[j].l];
}
printf("%d\n",ans);
}
return 0;
}
要不先看看莫队的题。。。
BZOJ-2038
听说比较像是模板题所以就先下刀了QAQ。。。调了好久发现第一个sort的位置放到了求p数组的前面,然后就愉快GG了
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#define NN 50010
#define LL long long
using namespace std;
struct REC
{
int l,r,num;
LL x,y;
}a[NN];
int c[NN],p[NN],block,n,m;
LL s[NN],ans=0;
bool cmp1(REC xx,REC yy)
{ if (p[xx.l]==p[yy.l]) return xx.r<yy.r; else return xx.l<yy.l;}
LL GCD(LL xx,LL yy)
{
while (xx) xx^=yy^=xx^=yy%=xx;
return yy;
}
bool cmp2(REC xx,REC yy) {return xx.num<yy.num;}
LL sqr(LL xx) { return xx*xx;}
void update(int xx,int yy)
{
ans-=sqr(s[c[xx]]);
s[c[xx]]+=yy;
ans+=sqr(s[c[xx]]);
}
int main()
{
scanf("%d%d",&n,&m);
for (int i=1;i<=n;++i) scanf("%d",&c[i]);
for (int i=1;i<=m;++i) scanf("%d%d",&a[i].l,&a[i].r),a[i].num=i;
block=int(sqrt(n));
for (int i=1;i<=n;++i) p[i]=(i-1)/block+1;
sort(a+1,a+1+m,cmp1);
for (int i=1,l=1,r=0;i<=m;++i)
{
for (;r<a[i].r;++r) update(r+1,1);
for (;r>a[i].r;--r) update(r,-1);
for (;l<a[i].l;++l) update(l,-1);
for (;l>a[i].l;--l) update(l-1,1);
if (a[i].l==a[i].r) { a[i].x=0ll,a[i].y=1ll; continue;}
a[i].x=ans-(a[i].r-a[i].l+1),a[i].y=(LL)(a[i].r-a[i].l+1)*(a[i].r-a[i].l);
LL tmp=GCD(a[i].x,a[i].y);
a[i].x/=tmp,a[i].y/=tmp;
}
sort(a+1,a+1+m,cmp2);
for (int i=1;i<=m;++i) printf("%lld/%lld\n",a[i].x,a[i].y);
return 0;
}
BZOJ-1878
突然发现怎么边上的人都在艹这道题。。。第一眼看过去好像莫队啊(中毒。。。)题解告诉我这是树状数组?然后再差点弃疗的时候。。。窝还是写了这道题
其实把上面的GCD求最简分数的地方去掉。。。再把update里面的东东改一改。。。就1A辣O(∩_∩)O~~
#include <iostream>
#include <cstdio>
#include <cstdlib>
#include <cmath>
#include <cstring>
#include <algorithm>
#define NN 50010
#define MM 200010
#define CC 1000010
#define LL long long
using namespace std;
struct REC
{
int l,r,x,num;
}a[MM];
int c[NN],s[CC],ans,p[NN],n,m,block;
void update(int xx,int yy)
{
if (s[c[xx]]) ans--;
s[c[xx]]+=yy;
if (s[c[xx]]) ans++;
}
bool cmp1(REC xx,REC yy)
{ if (p[xx.l]==p[yy.l]) return xx.r<yy.r; else return xx.l<yy.l;}
bool cmp2(REC xx,REC yy) {return xx.num<yy.num;}
int main()
{
scanf("%d",&n);
for (int i=1;i<=n;++i) scanf("%d",&c[i]);
scanf("%d",&m);
for (int i=1;i<=m;++i) scanf("%d%d",&a[i].l,&a[i].r),a[i].num=i;
block=int(sqrt(n));
for (int i=1;i<=n;++i) p[i]=(i-1)/block+1;
sort(a+1,a+1+m,cmp1);
for (int i=1,l=1,r=0;i<=m;++i)
{
for (;r<a[i].r;r++) update(r+1,1);
for (;r>a[i].r;r--) update(r,-1);
for (;l<a[i].l;l++) update(l,-1);
for (;l>a[i].l;l--) update(l-1,1);
a[i].x=ans;
}
sort(a+1,a+1+m,cmp2);
for (int i=1;i<=m;++i) printf("%d\n",a[i].x);
return 0;
}
所以本人好渣啊。。。。
【特长(zhǎng)生来了,本宝宝就被塔萌碾压惹
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