非官の题解 | 萃香的请柬

给几个结论：

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核心代码：

int p = lower_bound(fib + 1,fib + 101,k) - fib - 1;
return t[p] + f(k - fib[p]);

fib[1] = fib[2] = 1;
t[1] = t[2] = 1;
for(int i = 3;i <= 100;i++) fib[i] = fib[i - 1] + fib[i - 2];
for(int i = 3;i <= 100;i++) t[i] = f(fib[i]);

cout << f(r) - f(l - 1) << "\n";

完整代码根据思路自己写（