First One

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)

Total Submission(s): 142    Accepted Submission(s): 37

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

 

Source

 

因为下取整log（sum）的值是非常小的。

能够枚举每一个位置为開始位置，然后枚举每一个log（sum）仅仅需36*n的。

中间j的累加和

推公式就可以。

可是找log值同样的区间，须要用log(sum)*n的复杂度预处理出来，假设每次二分位置会超时。

#include
     
      #include
      
       #include
       
        #include
        
         #include
         
          using namespace std;#define ll long long#define maxn 100007ll num[maxn];ll pos[maxn][36];int main(){    int t;    scanf("%d",&t);    while(t--){        int n;        scanf("%d",&n);        for(int i = 0;i < n; i++){            scanf("%I64d",&num[i]);        }        num[n] = 0;        for(ll i = 0;i < 36; i++){            ll di = 1LL<<(i+1);            ll su = num[0];            int p = 0;            for(int j = 0;j < n; j++){                if(j) su -= num[j-1];                while(su < di && p < n){                    su += num[++p];                }                pos[j][i] = p;            }        }        ll ans = 0,res;        for(int i = 0;i < n; i++){            ll p = i,q;            for(int j = 0;j < 36 ;j ++){                q = pos[i][j];                res = (j+1)*((i+1)*(q-p)+(p+q+1)*(q-p)/2);                ans += res;                p = q;            }        }        printf("%I64d\n",ans);    }    return 0;}