• 分治思想


    Jon fought bravely to rescue the wildlings who were attacked by the white-walkers at Hardhome. On his arrival, Sam tells him that he wants to go to Oldtown to train at the Citadel to become a maester, so he can return and take the deceased Aemon's place as maester of Castle Black. Jon agrees to Sam's proposal and Sam sets off his journey to the Citadel. However becoming a trainee at the Citadel is not a cakewalk and hence the maesters at the Citadel gave Sam a problem to test his eligibility.

    Initially Sam has a list with a single element n. Then he has to perform certain operations on this list. In each operation Sam must remove any element x, such that x > 1, from the list and insert at the same position , , sequentially. He must continue with these operations until all the elements in the list are either 0 or 1.

    Now the masters want the total number of 1s in the range l to r (1-indexed). Sam wants to become a maester but unfortunately he cannot solve this problem. Can you help Sam to pass the eligibility test?

    Input

    The first line contains three integers n, l, r (0 ≤ n < 250, 0 ≤ r - l ≤ 105, r ≥ 1, l ≥ 1) – initial element and the range l to r.

    It is guaranteed that r is not greater than the length of the final list.

    Output

    Output the total number of 1s in the range l to r in the final sequence.

    Example
    Input
    7 2 5
    Output
    4
    Input
    10 3 10
    Output
    5
    Note

    Consider first example:

    Elements on positions from 2-nd to 5-th in list is [1, 1, 1, 1]. The number of ones is 4.

    For the second example:

    Elements on positions from 3-rd to 10-th in list is [1, 1, 1, 0, 1, 0, 1, 0]. The number of ones is 5.

    题目分析 : 给你一个 n ,将 n 分成 0 和 1,询问你给定区间内的 1 的个数

    思路分析 : 是一个分治的思想,每次只是查询所给区间的 1 的个数

    代码示例 :

    #define ll long long
    const int maxn = 1e6+5;
    const double pi = acos(-1.0);
    const int inf = 0x3f3f3f3f;
    ll n, l, r;
    
    ll fun(ll x){
        if (x <= 1) return 1;
        else return fun(x/2)*2+1;
    }
    
    ll dfs(ll n, ll l, ll r, ll a, ll b){ 
        if (l > b || r < a) return 0;
        if (n == 0) return 0;
        if (n == 1) return 1;
        ll sum = 0;
        ll mid = (a + b) >> 1;
        sum += dfs(n/2, l, r, a, mid-1);
        sum += dfs(n%2, l, r, mid, mid);
        sum += dfs(n/2, l, r, mid+1, b);
        return sum;
    }
    
    int main() {
        //freopen("in.txt", "r", stdin);
        //freopen("out.txt", "w", stdout);
        cin >> n >> l >> r;
        ll len = fun(n);
        printf("%lld
    ", dfs(n, l, r, 1ll, len));
        return 0;
    }
    
    东北日出西边雨 道是无情却有情
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  • 原文地址:https://www.cnblogs.com/ccut-ry/p/8452592.html
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