bjfu1277 简单递归

比较简单的递归问题。对于第k时刻的图形，可以平均分成四块，左上，右上，左下这三块的图形是一模一样的，右下的那一块不包含红毛僵尸，所以把那三块里的加起来就是结果了。

/*
 * Author    : ben
 */
#include <cstdio>
#include <cstdlib>
#include <cstring>
#include <cmath>
#include <ctime>
#include <iostream>
#include <algorithm>
#include <queue>
#include <set>
#include <map>
#include <stack>
#include <string>
#include <vector>
#include <deque>
#include <list>
#include <functional>
#include <numeric>
#include <cctype>
using namespace std;
typedef long long LL;
LL three[40];

LL work(int k, int a, int b) {
    int side = 1 << (k - 1);
    if (a > side || b < 1) {
        return 0;
    }
    if (a <= 1 && b >= side) {
        return three[k - 1];
    }
    int aa = a - (1 << (k - 2));
    int bb = b - (1 << (k - 2));
    return 2 * work(k - 1, a, b) + work(k - 1, aa, bb);
}

int main() {
    int T, a, b, k;
    three[0] = 1;
    for (int i = 1; i <= 30; i++) {
        three[i] = three[i - 1] * 3;
    }
    scanf("%d", &T);
    for (int t = 1; t <= T; t++) {
        scanf("%d%d%d", &k, &a, &b);
        printf("Case %d: %lld
", t, work(k, a, b));
    }
    return 0;
}