First One HDU - 5358

原题链接
考察:双指针
思路:
  很明显可以枚举\(log_2sum(i,j)\)的值,然后枚举左端点求右端点的区间,用二分TLE到我整个人都麻了,看题解是用双指针...
  我自己想的是用枚举右端点,二分求左端点区间,也是TLE...

Code

#include <iostream>
#include <cstring>
#include <algorithm>
using namespace std;
typedef long long LL;
const int N = 100010,M = 35;
int n;
LL sum[N],logs[M];
int main()
{
    int T;
    scanf("%d", &T);
	for(int i=1;i<M;i++) logs[i] = 1ll<<i;
    while(T--)
    {
        scanf("%d", &n);
        for (int i = 1; i <= n;i++)
        {
            scanf("%lld", &sum[i]);
            sum[i] += sum[i - 1];
        }
        LL res = 0;
        for(int j=0;j<35;j++)
        {
        	if(sum[n]<logs[j]) break;
        	int l = 1,r = 0;
        	for(int i=1;i<=n;i++)
        	{
        		if(sum[i-1]+logs[j]>sum[n]) break;
//        		int l = lower_bound(sum+i,sum+n+1,sum[i-1]+logs[j])-sum;
//       		int r = lower_bound(sum+i,sum+n+1,sum[i-1]+logs[j+1])-sum;
                l = max(l,i);
                while(l<=n&&sum[l]-sum[i-1]<logs[j]) l++;
                r = max(l-1,r);
                while(r+1<=n&&sum[r+1]-sum[i-1]<logs[j+1]) r++;
				if(l>r) continue;
				res+=((LL)(r-l+1)*(l+r)/2+(LL)(r-l+1)*i)*(j+1); 
			}
		}
        printf("%lld\n", res);
    }
    return 0;
}