【链接】

【题意】

在这里输入题意

【题解】

规律+递归题

f[k][i] k时刻前i行的红气球个数

i<=2^(k-1)

f[k][i] = 2*f[k-1][i];

i>2^(k-1)        f[k][i] = 2*c[k-1] + f[k-1][i-2^(k-1)];c[k]表示k时刻红气球个数显然k时刻有3^k个红气球

【代码】

#include 
    
     #define ll long longusing namespace std;const int N = 30;int k,a,b;ll two[N+10],three[N+10];ll f(int k,int i){    if (i==0) return 0;    if (k==0) return 1;    if (i <= (two[k-1]))        return 2*f(k-1,i);    else        return 2*three[k-1] + f(k-1,i-two[k-1]);}int main(){    #ifdef LOCAL_DEFINE        freopen("rush_in.txt", "r", stdin);    #endif    ios::sync_with_stdio(0),cin.tie(0);    int T;    two[0] = three[0] = 1;    for (int i = 1;i <= N;i++)        two[i] = two[i-1]*2,three[i] = three[i-1]*3;    int kase = 0;    cin >> T;    while (T--){        cin >> k >> a >> b;        cout <<"Case "<<++kase<<": "<< f(k,b) - f(k,a-1)<