hdu5358 First One

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
这题题意容易懂，就是求和，其中(⌊log2S(i,j)⌋+1)的意思就是S(i,j)化成二进制后的比特位个数，因为S(i,j)不超过10^10，所以比特位不会超过35个。我们可以先初始化b[],记录比特位为i的所有数中的最后一个数2^i-1，用sum[i]把从1到i的总和记录下来，然后用35个指针pt[i]记录以i为起点的最大下标k满足sum[k]-sum[i-1]<=b[j]。最后注意要用G++交，C++会超时。。
#include<iostream>
#include<stdio.h>
#include<stdlib.h>
#include<string.h>
#include<math.h>
#include<vector>
#include<map>
#include<set>
#include<queue>
#include<stack>
#include<string>
#include<algorithm>
using namespace std;
#define ll long long
#define maxn 100060
ll b[50],sum[maxn];//b[1]=2^i-1
ll a[maxn];
int pt[44];//指针
void init()
{
	int i,j;
	b[0]=-1;
	b[1]=1;
	for(i=2;i<=35;i++){
		b[i]=(1LL<<i)-1; //也可以是b[i]=((ll)1<<i)-1;,但不加的话会爆int 
	}
}
int main()
{
	int n,m,i,j,T,len;
	ll ans;
	init();
	scanf("%d",&T);
	while(T--)
	{
		scanf("%d",&n);
		sum[0]=0;ans=0;
		for(i=1;i<=n;i++){
			scanf("%lld",&a[i]);
			sum[i]=sum[i-1]+a[i];
		}
		for(i=1;i<=35;i++)pt[i]=0;
		for(i=1;i<=n;i++){
			pt[0]=i-1;
			for(j=1;j<=34;j++){
				while(sum[pt[j]+1]-sum[i-1]<=b[j] && pt[j]<n){//如果a>b，那么pt[a]一定大于等于pt[b] 
					pt[j]++;
				}
				//if(sum[pt[j]]-sum[i-1]>b[j-1] && sum[pt[j]]-sum[i-1]<=b[j] && pt[j]>=i ){ 这一句可以不用写 
					len=(pt[j]-pt[j-1]);
					ans+=(ll)j*len*i;
					ans+=(ll)j*len*(pt[j-1]+1+pt[j])/2;
				//}
			}
		}
		printf("%lld\n",ans);
	}
	return 0;
}