PC/UVA 110101/100
水题。数据范围大。需要预处理。之前处理过的+处理到这个地方的
//author: CHC //First Edit Time: 2014-01-10 18:30 //Last Edit Time: 2014-01-10 18:30 //Filename:1.cpp #include <iostream> #include <cstdio> #include <string.h> #include <queue> #include <algorithm> using namespace std; long long cs[1000001]={ 0 }; int main() { for(int j=1;j<1000001;j++) { int i=1; long long n=j; while(n!=1) { if(n%2)n=n*3+1; else n/=2; if(n<j) { cs[j]=cs[n]+i; break; } ++i; } if(!cs[j]) cs[j]=i; } int a,b,c,d; while(~scanf("%d%d",&c,&d)) { a=c,b=d; if(a>b)a^=b^=a^=b; int maxn=cs[a],maxi=a; for(int i=a+1;i<=b;i++) if(cs[i]>maxn) { maxn=cs[i]; maxi=i; } printf("%d %d %d\n",c,d,maxn); } return 0; }
|
||||||
Consider the following algorithm to generate a sequence of numbers. Start with an integer n. If n is even, divide by 2. If n is odd, multiply by 3 and add 1. Repeat this process with the new value of n, terminating when n = 1. For example, the following sequence of numbers will be generated for n = 22:
For an input n, the cycle-length of n is the number of numbers generated up to and including the 1. In the example above, the cycle length of 22 is 16. Given any two numbers i and j, you are to determine the maximum cycle length over all numbers between i and j, including both endpoints.
1 10 100 200 201 210 900 1000
1 10 20 100 200 125 201 210 89 900 1000 174