?

• [1457] Sona

• 時間限制: 5000 ms　內存限制: 65535 K
• 問題描述
• Sona,?Maven of the Strings. Of cause, she can play the zither.

  ?

  Sona?can't speak but she can make fancy music. Her music can attack, heal, encourage and enchant.

  ?

  There're an ancient?score(樂譜). But because it's too long,?Sona?can't play it in a short moment. So?Sona?decide to just play a part of it and revise it.

  ?

  A score is composed of notes. There?are?109?kinds of notes and a score has105?notes at most.

  ?

  To diversify?Sona's own score, she have to select several parts of it. The energy of each part is calculated like that:

  ?

  Count the number of times that each notes appear. Sum each of the number of times' cube together. And the sum is the energy.

  ?

  You should help?Sona?to calculate out the energy of each part.

• 輸入
• This problem contains several cases. And this problem provides 2 seconds to run.
  The first line of each case is an integer N (1 ≤ N ≤ 10^5), indicates the number of notes.
  Then N numbers followed. Each number is a kind of note. (1 ≤ NOTE ≤ 10^9)
  Next line is an integer Q (1 ≤ Q ≤ 10^5), indicates the number of parts.
  Next Q parts followed. Each part contains 2 integers Li and Ri, indicates the left side of the part and the right side of the part.
• 輸出
• For each part, you should output the energy of that part.
• 樣例輸入

• 8
  1 1 3 1 3 1 3 3
  4
  1 8
  3 8
  5 6
  5 5
  

• 樣例輸出

• 128
  72
  2
  1

?

?

題目鏈接：NBUT 1457

莫隊算法就是能用n*sqrt（n）的時間來解決無修改區間查詢的一種辦法……，主要處理就是把詢問暫時不按照讀入順序而按照所在塊和右端點值為關鍵字進行排序，然后對這樣的一個暫時非正常的詢問順序進行計算回答，然后可以再次按照輸入順序排序或者用數組記錄答案后輸出。

可以參考這篇博客：莫隊算法經典例題

這題寫出立方公式就很容易發現a^3和（a+1）^3以及（a-1）^3的區別，然后就可以很方便地進行O（1）轉化，網上一些題解代碼都是暴力減掉當前的再加回更新的，即減掉a*a*a再加上(a+1)*(a+1)*(a+1)，感覺這樣寫不太好……，反正如果能快速得到[L-1,R]或[L,R+1]這樣的區間信息的話莫隊算法值得一試。這題用了輸入輸出外掛搞到2100+MS，一些人直接用立方加減的3300+MS

代碼：

#include<iostream>
#include<algorithm>
#include<cstdlib>
#include<sstream>
#include<cstring>
#include<bitset>
#include<cstdio>
#include<string>
#include<deque>
#include<stack>
#include<cmath>
#include<queue>
#include<set>
#include<map>
using namespace std;
#define INF 0x3f3f3f3f
#define CLR(x,y) memset(x,y,sizeof(x))
#define LC(x) (x<<1)
#define RC(x) ((x<<1)+1)
#define MID(x,y) ((x+y)>>1)
typedef pair<int,int> pii;
typedef __int64 LL;
const double PI=acos(-1.0);
const int N=100010;struct info
{int l;int r;LL energy;int id;
};
info node[N];
LL arr[N];
int unit,b[N];
LL cnt[N];
LL ans[N];
inline bool cmp(const info &a,const info &b)
{if(a.l/unit!=b.l/unit)return a.l/unit<b.l/unit;return a.r<b.r;
}void init()
{CLR(arr,0);CLR(cnt,0);
}
inline void update(int pos,int n,LL &temp)
{LL &val=arr[pos];if(n==1)temp=temp+3*cnt[val]*cnt[val]+3*cnt[val]+1;elsetemp=temp-3*cnt[val]*cnt[val]+3*cnt[val]-1;cnt[val]+=n;	
}
int Scan()
{int res=0,ch,flag=0;if((ch=getchar())=='-')flag=1;else if(ch>='0'&&ch<='9')res=ch-'0';while((ch=getchar())>='0'&&ch<='9')res=res*10+ch-'0';return flag?-res:res;
}
void Out(LL a)
{if(a>9)Out(a/10LL);putchar(a%10LL+'0');
}
int main(void)
{int n,i,j,k,L,R;while (~scanf("%d",&n)){for (i=1; i<=n; ++i){arr[i]=Scan();b[i]=arr[i];}sort(b+1,b+1+n);int m=unique(b+1,b+1+n)-b-1;for (i=1; i<=n; ++i)arr[i]=lower_bound(b+1,b+1+m,arr[i])-b;unit=(int)sqrt(1.0*n);scanf("%d",&k);for (i=0; i<k; ++i){node[i].l=Scan();node[i].r=Scan();node[i].id=i;}sort(node,node+k,cmp);L=node[0].l;R=node[0].l-1;LL temp=0;for (i=0; i<k; ++i){while (L>node[i].l)update(--L,1,temp);while (R<node[i].r)update(++R,1,temp);while (L<node[i].l)update(L++,-1,temp);while (R>node[i].r)update(R--,-1,temp);ans[node[i].id]=temp;}for (i=0; i<k; ++i){Out(ans[i]);putchar('\n');}init();}return 0;
}

轉載于:https://www.cnblogs.com/Blackops/p/5766255.html

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