类似Fibonacci数列,记得这种题用递归做你就死翘翘了(复杂度O(2^N))。好歹也应该设一个数组再递推的(复杂度O(N))关于Fibonacci数列,TJU1248的难度……偶现在还是tle……努力思索中,哈哈~~帖1005的代码:#include<iostream>using namespace std;int Cows(int Years);int main(){ int yrs,cow; while(cin>>yrs) { cow=Cows(yrs); cout<<cow<<endl; } return 0;}int Cows(int Years){ if(Years<4) return 1; else { int temp=4; int cow[]={1,1,1,1}; for(;temp<=Years;temp++) cow[temp%4]=cow[(temp-1)%4] + cow[(temp-3)%4]; return cow[Years%4]; }} 转载于:https://www.cnblogs.com/FancyMouse/articles/219502.html