hdu 5358 First One 2015多校联合训练赛#6 枚举

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

2015 Multi-University Training Contest 6

 

由于下取整log（sum）的值是很小的。可以枚举每个位置为开始位置，然后枚举每个log（sum）只需36*n的。中间j的累加和

推公式即可。

但是找log值相同的区间，需要用log(sum)*n的复杂度预处理出来，如果每次二分位置会超时。

#include<iostream>
#include<cstdio>
#include<cstring>
#include<algorithm>
#include<vector>
using namespace std;
#define ll long long
#define maxn 100007
ll 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;
}