UVA 12627 Erratic Expansion（奇怪的气球膨胀）（递归）

题意：问k小时后，第A~B行一共有多少个红气球。

分析：观察图可发现，k小时后，图中最下面cur行的红气球个数满足下式：

(1)当cur <= POW[k - 1]时,

dfs(k, cur) = dfs(k - 1, cur);

(2)当cur > POW[k - 1]时，

dfs(k - 1, cur) = 2 * dfs(k - 1, cur - POW[k - 1]) + tot[k - 1];

其中，POW[k - 1]为2^(k  - 1)，tot[k - 1]为k-1小时后图中的红气球总数，为3^(k - 1)。

#pragma comment(linker, "/STACK:102400000, 102400000")
#include<cstdio>
#include<cstring>
#include<cstdlib>
#include<cctype>
#include<cmath>
#include<iostream>
#include<sstream>
#include<iterator>
#include<algorithm>
#include<string>
#include<vector>
#include<set>
#include<map>
#include<stack>
#include<deque>
#include<queue>
#include<list>
#define Min(a, b) ((a < b) ? a : b)
#define Max(a, b) ((a < b) ? b : a)
typedef long long LL;
typedef unsigned long long ULL;
const int INT_INF = 0x3f3f3f3f;
const int INT_M_INF = 0x7f7f7f7f;
const LL LL_INF = 0x3f3f3f3f3f3f3f3f;
const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;
const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};
const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};
const int MOD = 1e9 + 7;
const double pi = acos(-1.0);
const double eps = 1e-8;
const int MAXN = 10000 + 10;
const int MAXT = 10000 + 10;
using namespace std;
LL tot[35];
LL POW[35];//记录2的次方
void init(){
    tot[0] = 1;
    POW[0] = 1;
    for(int i = 1; i <= 30; ++i){
        tot[i] = 3 * tot[i - 1];
        POW[i] = 2 * POW[i - 1];
    }
}
LL dfs(int k, int cur){
    if(cur == 0) return 0;
    if(k == 0) return 1;
    if(cur <= POW[k - 1]){
        return dfs(k - 1, cur);
    }
    else{
        return 2 * dfs(k - 1, cur - POW[k - 1]) + tot[k - 1];
    }
}
int main(){
    int T;
    scanf("%d", &T);
    int kase = 0;
    init();
    while(T--){
        int k, a, b;
        scanf("%d%d%d", &k, &a, &b);
        printf("Case %d: %lld\n", ++kase, dfs(k, POW[k] - a + 1) - dfs(k, POW[k] - b));
    }
    return 0;
}