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:
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).
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<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;
}