[UVA][遞迴] 12627 - Erratic Expansion
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Erratic Expansion Input: Standard Input Output: Standard Output |
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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 0th, 1st, 2nd and 3rd hour are depicted in the following diagram.
As you can see, a red balloon in the cell (i, j) (that is ith row and jth 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 Kth 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 ≤ 2K.
Output
For each case, output the case number followed by the total number of red balloons in rows [A, B] after Kth hour.
Sample Input Output for Sample Input
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3 0 1 1 3 1 8 3 3 7
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Case 1: 1 Case 2: 27 Case 3: 14 |
Problemsetter: Sohel Hafiz, Special Thanks: Md. Mahbubul Hasan
題目解法：
還好這題 column, row 怎麼擺放並不重要, 剛好對稱不影響。
這題就遞迴解決它, 不過要預算一個 (2^k)*(2^k) 如果全選的個數是多少, 得到了 3^k 個紅色氣球。
#include <stdio.h>
long long I[50];
long long dfs(int k, int i, int j) {
if(k == 0) return 1;
long long ret = 0;
int w = 1<<(k-1);
if(i == 1 && j == w*2)
return I[k];
if(w >= j)
return dfs(k-1, i, j)*2;
if(i > w)
return dfs(k-1, i-w, j-w);
return dfs(k-1, i, w)*2 + dfs(k-1, 1, j-w);
}
int main() {
I[0] = 1;
for(int i = 1; i < 50; i++)
I[i] = I[i-1]*3;
int testcase, cases = 0;
scanf("%d", &testcase);
while(testcase--) {
int k, i, j;
scanf("%d %d %d", &k, &i, &j);
printf("Case %d: %lld\n", ++cases, dfs(k, i, j));
}
return 0;
}