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 = ;

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

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

     int con = ;

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

     return ;
 }