UVA12627-Erratic Expansion（递归）

Problem UVA12627-Erratic Expansion

Accept: 465  Submit: 2487
Time Limit: 3000 mSec

Problem Description

Input

The first line of input is an integer T (T < 1000) that indicates the number of test cases. Each case contains 3 integers K, A and B. The meanings of these variables are mentioned above. K will be in the range [0,30] and 1 ≤ A ≤ B ≤ 2K.

 Output

For each case, output the case number followed by the total number of red balloons in rows [A,B] after K-th hour.

 

 Sample Input

3

0 1 1

3 1 8

3 3 7

 

 Sample Output

Case 1: 1

Case 2: 27

Case 3: 14

题解：本来想找规律，但是纯粹找规律不太好找，参考了一下lrj的思路，设g(k,i)为k小时后第i行及以下的红气球个数，这样递归方程就很好推了（详见代码），红气球的个数很显然是3^k，还有一个注意的点是，举个例子，k=2，则此时最多4行，如果求第5行及以下的气球数，直接返回0即可。

#include <bits/stdc++.h>

using namespace std;
typedef long long LL;

const int maxn = 50;

int k, a, b;
int two_pow[maxn];
LL tri_pow[maxn];

LL g(int k, int i) {
    if (i > two_pow[k]) return 0LL;
    if (k == 0) return 1LL;
    if (i > two_pow[k - 1]) {
        return g(k - 1, i - two_pow[k - 1]);
    }
    else {
        return 2 * g(k - 1, i) + tri_pow[k - 1];
    }
}

int main()
{
    //freopen("input.txt", "r", stdin);
    int iCase;
    scanf("%d", &iCase);
    two_pow[0] = 1;
    tri_pow[0] = 1LL;
    for (int i = 1; i <= 30; i++) {
        two_pow[i] = two_pow[i - 1] * 2;
        tri_pow[i] = tri_pow[i - 1] * 3;
    }

    int con = 1;

    while (iCase--) {
        scanf("%d%d%d", &k, &a, &b);
        printf("Case %d: %lld
", con++, g(k, a) - g(k, b + 1));
    }

    return 0;
}