hdu 5358 First One

　　题目链接：hdu 5358

　　思路不难理解，就是个尺取法而已，floor(log₂X) + 1 就是求 X 的二进制表示的位数，对于题目来说这个值最多只是 30+，从这里入手开始枚举，运用尺取法可以达到 O(n) 的复杂度，具体百度之，按照这个思路写的代码 wa 了无数遍，一下午又这样没了，唉，好无语啊~~让我吃尽苦头的一道题，先记录下来，有空再慢慢琢磨。已AC代码：

#include<cstdio>
#include<cstring>
#include<algorithm>
using namespace std;
typedef long long ll;
const int N = 100005;

int n, a[N];
ll sum[N];
ll p2[40] = {1, };
inline void init(int n = 35) {
    for(int i = 1; i <= n; ++i)
        p2[i] = p2[i - 1] * 2;
}

inline ll func(ll L, ll r1, ll r2) {
    return (L + r1 + L + r2) * (r2 - r1 + 1) / 2;
}

ll cal(int k) {
    ll res = 0;
    ll low = (k == 1 ? 0: p2[k - 1]), up = p2[k] - 1;
    int r1 = 1, r2 = 0;
    for(int L = 1; L <= n; ++L) {
        r1 = max(r1, L);
        while(r1 <= n && (sum[r1] - sum[L - 1]) < low)   ++r1;
        if(r1 > n || sum[r1] - sum[L - 1] > up)   continue;

        r2 = max(r2, r1 - 1);
        while(r2 + 1 <= n && sum[r2 + 1] - sum[L - 1] >= low
              && sum[r2 + 1] - sum[L - 1] <= up)    ++r2;

        if(r1 > r2)   continue;
        res += func(L, r1, r2);
    }
    return res * k;
}

int main() {
    int t;
    init();
    scanf("%d",&t);
    while(t--) {
        scanf("%d",&n);
        for(int i = 1; i <= n; ++i) {
            scanf("%d", a + i);
            sum[i] = sum[i - 1] + a[i];
        }
        ll ans = 0;
        for(int i = 1; i <= 35; ++i)
            ans += cal(i);
        printf("%I64d\n",ans);
    }
    return 0;
}