uva12627 Erratic Expansion (递归)

Erratic Expansion

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Piotr found a magical box in heaven. Its magic power is that if you place any red balloon inside it then, after one hour, it will multiply to form 3 red and 1 blue colored balloons. Then in the next hour, each of the red balloons will multiply in the same fashion, but the blue one will multiply to form 4 blue balloons. This trend will continue indefinitely.
The arrangements of the balloons after the 0-th, 1-st, 2-nd and 3-rd hour are depicted in the following diagram.

As you can see, a red balloon in the cell (i, j) (that is i-th row and j-th column) will multiply to produce 3 red balloons in the cells (i ∗ 2 − 1, j ∗ 2 − 1), (i ∗ 2 − 1, j ∗ 2), (i ∗ 2, j ∗ 2 − 1) and a blue balloon in the cell (i ∗ 2, j ∗ 2). Whereas, a blue balloon in the cell (i, j) will multiply to produce 4 blue balloons in the cells (i ∗ 2 − 1, j ∗ 2 − 1), (i ∗ 2 − 1, j ∗ 2), (i ∗ 2, j ∗ 2 − 1) and (i ∗ 2, j ∗ 2). The grid size doubles (in both the direction) after every hour in order to accommodate the extra balloons.
In this problem, Piotr is only interested in the count of the red balloons; more specifically, he would like to know the total number of red balloons in all the rows from A to B after K-th hour.
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 ≤ 2 K.
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

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这题只要求我们看红色气球的数量，我们把每行的红色气球数量竖着写下来，很容易就可以发现这张图形的变化规律，即每次图形上方复制出一个图形，上面每行的数量是下面每行的两倍，原来的图形的数量不变。这里需要利用递归的思想来写。

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#include <cstdio>
#include <iostream>
using namespace std;
int k;
int Pow(int a,int b)
{
    int base =a;
    int ans = 1;
    while(b!=0)
    {
        if(b&1)
            ans*=base;
        b>>=1;
        base*=base;
    }
    return ans;
}
long long c(int k)
{
    if(k==0)return 1;
    else
        return 3*c(k-1);
}
long long f(int k,int i)
{
    if(i<=0) return 0;
    if(k==0) return 1;
    if(i>=Pow(2,k-1)) return (f(k-1,i-Pow(2,k-1))+2*c(k-1));
    else return 2*f(k-1,i);
}
int main()
{

    int t;
    cin>>t;
    int k,a,b;
    int kase=0;
    while(t--)
    {
        scanf("%d %d %d",&k,&a,&b);
        long long ans =f(k,b)-f(k,a-1);
        printf("Case %d: %lld\n",++kase,ans);
    }

    return 0;
}