版权声明:本文为博主原创文章,未经博主允许不得转载。 https://blog.csdn.net/jirryzhang/article/details/82502902
递推公式如下:
n<=m时
v[n]=2<<(n-2)
n>m时
v[n]=2*v[n-1]-v[n-1-m]
输入n超过10^18,考虑是维护一个供迭代的双向队列做动态规划,更新尾部第i级台阶的跳法数,删除头部第i-m-1级台阶的跳法数,直到算出第n级的解。
#include<iostream>
#include<deque>
using namespace std;
int main()
{
int m;
long long n;
int top,end;
while(cin>>n>>m){
if(n<2)
cout<<1<<endl;
else if(n<=m)
cout<<(2 << (n - 2)) % 10007<<endl;
else {
deque<int> d;
d.push_back(1);
d.push_back(1);
for (auto i = 2; i <= m; ++i) {
d.push_back((2 << (i - 2)) % 10007);
}
for (auto i = m + 1; i <= n; ++i) {
top = d.back();
end = d.front();
d.pop_front();
d.push_back((2 * top - end) % 10007);
}
top = d.back();
while (top < 0)
top += 10007;
cout << top << endl;
}
}
return 0;
}