hdu-5358 First One(尺取法)

题目链接：

First One

Time Limit: 4000/2000 MS (Java/Others)    

Memory Limit: 131072/131072 K (Java/Others)

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

 

题意：

 

求这个sum啦啦啦;

 

思路：

 

还是得用尺取法,我二分找两个端点,tle了,看来常数卡的好紧啊啊啊;

 

 

AC代码：

 

/*    5358    1638MS    3120K    1119 B    G++    2014300227*/
#include <bits/stdc++.h>
using namespace std;
const int N=1e5+4;
typedef long long ll;
int n;
ll a[N],sum[N];
int main()
{
    int t;
    scanf("%d",&t);
    while(t--)
    {
        scanf("%d",&n);
        for(int i=1;i<=n;i++)
        {
            scanf("%lld",&a[i]);
        }
        sum[0]=0;
        a[n+1]=0;
        for(int i=1;i<=n+1;i++)
        {
            sum[i]=sum[i-1]+a[i];
        }
        ll ans=0,num=0;
        ll L=0,R=1,len;
        for(int i=0;i<=33;i++)
        {
            num=0;
            int l=1,r=1;
            for(int j=1;j<=n;j++)//左端点为j,l和r为右端点可取的区间;并不是l表示左端点,r表示右端点;所以复杂度才降了下来;
            {
                l=max(j,l);
                while(l<=n&&sum[l]-sum[j-1]<L)l++;
                r=max(r,l);
                while(r<=n&&sum[r]-sum[j-1]<=R&&sum[r]-sum[j-1]>=L)r++;
                len=r-l;
                num+=len*(ll)j+len*(l+r-1)/2;
            }
            L=R+1;
            R=2*L-1;
            ans+=num*(ll)(i+1);
        }

        printf("%lld
",ans);
    }
    return 0;
}