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 = 100010;
LL p[38] = {1},sum[maxn];
void init() {
    for(int i = 1; i <= 36; ++i)
        p[i] = (p[i-1]<<1);
}
int main() {
    init();
    int kase,n;
    scanf("%d",&kase);
    while(kase--){
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i){
            scanf("%I64d",sum+i);
            sum[i] += sum[i-1];
        }
        LL ret = 0;
        for(int i = 1; i <= 35 && p[i] <= sum[n]; ++i){
            int L = 1,R = 0;
            LL tmp = 0;
            for(int j = 1; j <= n; ++j){
                while(L <= n && sum[L] - sum[j-1] < p[i]) ++L;
                while(R + 1 <= n && sum[R + 1] - sum[j-1] < p[i+1]) ++R;
                if(L <= R) tmp += (LL)j*(R - L + 1) + 1LL*(L + R)*(R - L + 1)/2;
            }
            ret += tmp*i;
        }
        for(int i = 1; i <= n; ++i)
            ret += LL(n + i)*(n - i + 1)/2 + LL(i)*(n - i + 1);
        printf("%I64d\n",ret);
    }
    return 0;
}