hdu5358 First One 尺取法 多校联合第六场

First One

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 1511    Accepted Submission(s): 457

Problem Description

soda has an integer array a1,a2,…,an . Let S(i,j) be the sum of ai,ai+1,…,aj . Now soda wants to know the value below:

∑i=1n∑j=in(⌊log2S(i,j)⌋+1)×(i+j)

Note: In this problem, you can consider log20 as 0.

 

Input

There are multiple test cases. The first line of input contains an integer T , indicating the number of test cases. For each test case:

The first line contains an integer n (1≤n≤105) , the number of integers in the array.
The next line contains n integers a1,a2,…,an (0≤ai≤105) .

 

Output

For each test case, output the value.

 

Sample Input

  
  

   
   1
2
1 1
  
  

 

Sample Output

  
  

   
   12
  
  

  计算上面给的那个公式。

  比赛的时候想了个方法，以每个位置作为终点，二分log每种取值对应的起点，这样的复杂度是O（NlogN*35）左右，但是超时。这道题时间卡的很紧，复杂度O（N*35）的都要运行1000多ms。

  枚举log不同取值的区间范围，在O（N）的复杂度下算出这个范围对应的所有i,j和。采用尺取法，假设以i为起始位置，对应[p1,p2]这个区间为终止位置，这些区间都是满足的，那么当i+1时，一定有终止位置p1'>=p1,p2'>=p2才可能满足。

#include<iostream>
#include<cstdio>
#include<string>
#include<cstring>
#include<vector>
#include<cmath>
#include<queue>
#include<stack>
#include<map>
#include<set>
#include<algorithm>
using namespace std;
typedef long long LL;

const LL MAXN=100010;
const LL INF=0x3f3f3f3f;
const LL SIGMA_SIZE=28;

int T,N;
int a[MAXN];
LL sum[MAXN];

//[l,r) i+j的和
LL solve(LL l,LL r){
    LL ret=0;
    int p1=1,p2=1;
    for(int i=1;i<=N;i++){
        if(p1<i) p1=i;
        if(p2<i) p2=i;
        while(p1<=N&&sum[p1]-sum[i-1]<l) p1++;
        while(p2<=N&&sum[p2]-sum[i-1]<r) p2++;
        p2--;
        if(p1<=p2&&sum[p1]-sum[i-1]>=l&&sum[p2]-sum[i-1]<r){
            LL n=p2-p1+1;
            ret+=i*n+p1*n+n*(n-1)/2;
        }
    }
    return ret;
}

int main(){
    freopen("in.txt","r",stdin);
    scanf("%d",&T);
    while(T--){
        scanf("%d",&N);
        sum[0]=0;
        for(int i=1;i<=N;i++){
            scanf("%d",&a[i]);
            sum[i]=sum[i-1]+a[i];
        }
        LL ans=0;
        ans+=solve(0,1);
        for(int i=0;i<35;i++){
            ans+=solve(1LL<<i,1LL<<(i+1))*(i+1);
        }
        printf("%I64d\n",ans);
    }
    return 0;
}