2015 Multi-University Training Contest 6 hdu 5358 First One

First One

Time Limit: 4000/2000 MS (Java/Others)    Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 881    Accepted Submission(s): 273

 soda has an integer array $a_1,a_2,\dots,a_n$. Let $S(i,j)$ be the sum of $a_i,a_i+1,\dots,a_j$. Now soda wants to know the value below:

 \[\sum_{i = 1}^{n}\sum_{j = i}^{n}(\lfloor \log_{2}{S(i,j)} \rfloor + 1)\times (i+j) \]

Note: In this problem, you can consider $\log_{2}{0}$ 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 $n (1 \geq n \leq 10^5)$,the number of integers in the array.

The next line contains n integers $0 \leq a_{i} \leq 10^{5}$

 

Output

For each test case, output the value.

 

Sample Input

1

2

1 1

 

Sample Output

12

 

Source

2015 Multi-University Training Contest 6 F-1006

解题：这道题目有意思，我们可以发现 $a_{i} \leq 10^{5}$ 这是一个信息，破题的关键.

 

$\log_{2}{sum}$大概会在什么范围呢？sum最多是 $10^{5} \times 10^{5} = 10^{10}$

也就是说$\log_{2}{sum} \leq 34$ 

 

才35，sum的特点是什么?很明显都是非负数，那么sum必须是递增的，单调的，F-100. 我们可以固定$log_{2}{S(i,j)}$ 后 固定左区间j,找出以j作为左区间，然后当然是找出最小的r 和 最大的 R

 

最小 最大？当然是这样的区间，该区间的和取对数是我们刚刚固定的那个数。区间可以表示成这样 $[j,r\dots R]$ 那么从j到r,j到 r+1,...,j 到R ，这些区间的和取对数都会等于同一个数。。

 

好了如何算$[j,r\dots R]$ 的下标和？很明显吗。。j r,j r+1, j r+2,..., j R.把左区间下标一起算了，右区间是个等差数列，求和。

 

我们在最后把那个表达式里面的1加上

同样的计算方法

 

 #include <bits/stdc++.h>
 using namespace std;
 typedef long long LL;
 const int maxn = ;
 LL p[] = {},sum[maxn];
 void init() {
     for(int i = ; i <= ; ++i)
         p[i] = (p[i-]<<);
 }
 int main() {
     init();
     int kase,n;
     scanf("%d",&kase);
     while(kase--){
         scanf("%d",&n);
         for(int i = ; i <= n; ++i){
             scanf("%I64d",sum+i);
             sum[i] += sum[i-];
         }
         LL ret = ;
         for(int i = ; i <=  && p[i] <= sum[n]; ++i){
             int L = ,R = ;
             LL tmp = ;
             for(int j = ; j <= n; ++j){
                 while(L <= n && sum[L] - sum[j-] < p[i]) ++L;
                 while(R +  <= n && sum[R + ] - sum[j-] < p[i+]) ++R;
                 if(L <= R) tmp += (LL)j*(R - L + ) + 1LL*(L + R)*(R - L + )/;
             }
             ret += tmp*i;
         }
         for(int i = ; i <= n; ++i)
             ret += LL(n + i)*(n - i + )/ + LL(i)*(n - i + );
         printf("%I64d\n",ret);
     }
     return ;
 }