2015多校1006.First One

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原文链接：http://www.cnblogs.com/get-an-AC-everyday/p/4710925.html

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#java

本文解析了一道算法竞赛题目，该题涉及整数数组的区间和与对数运算，通过枚举和尺取法求解复杂度问题。文章分享了解题思路及优化方案，包括将浮动区间和转化为固定值处理的方法。

First One

Time Limit: 4000/2000 MS (Java/Others) Memory Limit: 131072/131072 K (Java/Others)
Total Submission(s): 831 Accepted Submission(s): 253

Problem Description

soda has an integer array

\sum i = 1 n \sum j = i n (⌊ log 2 S (i, j) ⌋ + 1) \times (i + j)

Input

There are multiple test cases. The first line of input contains an integer

Output

For each test case, output the value.

Sample Input

1 2 1 1

Sample Output

Source

2015 Multi-University Training Contest 6

 1 #include<bits/stdc++.h>
 2 using namespace std;
 3 typedef long long ll ;
 4 const int M = 1e5 + 10 ;
 5 ll n ;
 6 ll a[M] ;
 7 ll ure[M] ;
 8 ll tot ;
 9 void solve () {
10         for (int k = 1 ; k < 40 ; k ++) {
11                 ll low = 1ll << k , sum = 0 ;
12                 for (int i = 1 , j = 0 ; i <= n ; i ++) {
13                         while (j <= n && sum < low) sum += a[++j] ;
14                         if (sum >= low) tot += i * (n-j+1) + ure[j] ;
15                         else break ;
16                         sum -= a[i] ;
17                 }
18         }
19         printf ("%I64d\n" , tot) ;
20 }
21 
22 int main () {
23         int T ;
24         scanf ("%d" , &T ) ;
25         while (T --) {
26                 scanf ("%I64d" , &n) ;
27                 for (int i = 1 ; i <= n ; i ++) scanf ("%I64d" , &a[i]) ;
28 
29                 ure[n+1] = 0 ;
30                 tot = 0 ;
31                 for (int i = n ; i >= 1 ; i --) {
32                         ure[i] = ure[i+1] + i ;
33                         tot += i * (n-i+1) + ure[i] ;
34                 }
35                 
36                 solve () ;
37         }
38         return 0 ;
39 }