ACM--次数最多--HDOJ 1032--The 3n + 1 problem--水(数据太弱)
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HDOJ题目地址:传送门
The 3n + 1 problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 34456 Accepted Submission(s): 12454
Problem Description
Problems in Computer Science are often classified as belonging to a certain class of problems (e.g., NP, Unsolvable, Recursive). In this problem you will be analyzing a property of an algorithm whose classification is not known for all possible inputs.
Consider the following algorithm:
1. input n
2. print n
3. if n = 1 then STOP
4. if n is odd then n <- 3n + 1
5. else n <- n / 2
6. GOTO 2
Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)
Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.
Consider the following algorithm:
1. input n
2. print n
3. if n = 1 then STOP
4. if n is odd then n <- 3n + 1
5. else n <- n / 2
6. GOTO 2
Given the input 22, the following sequence of numbers will be printed 22 11 34 17 52 26 13 40 20 10 5 16 8 4 2 1
It is conjectured that the algorithm above will terminate (when a 1 is printed) for any integral input value. Despite the simplicity of the algorithm, it is unknown whether this conjecture is true. It has been verified, however, for all integers n such that 0 < n < 1,000,000 (and, in fact, for many more numbers than this.)
Given an input n, it is possible to determine the number of numbers printed (including the 1). For a given n this is called the cycle-length of n. In the example above, the cycle length of 22 is 16.
For any two numbers i and j you are to determine the maximum cycle length over all numbers between i and j.
Input
The input will consist of a series of pairs of integers i and j, one pair of integers per line. All integers will be less than 1,000,000 and greater than 0.
You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.
You can assume that no opperation overflows a 32-bit integer.
You should process all pairs of integers and for each pair determine the maximum cycle length over all integers between and including i and j.
You can assume that no opperation overflows a 32-bit integer.
Output
For each pair of input integers i and j you should output i, j, and the maximum cycle length for integers between and including i and j. These three numbers should be separated by at least one space with all three numbers on one line and with one line of output for each line of input. The integers i and j must appear in the output in the same order in which they appeared in the input and should be followed by the maximum cycle length (on the same line).
Sample Input
1 10100 200201 210900 1000
Sample Output
1 10 20100 200 125201 210 89900 1000 174
题意:
题的大意是输入两个数(注意,这里没有说第一个数一定要小与第二个数),然后对这两个数之间的所有整数包括端点的数,进行一种规律运算,并求出运算的次数,比较所有的数规律运算次数,输出最大的.
题的大意是输入两个数(注意,这里没有说第一个数一定要小与第二个数),然后对这两个数之间的所有整数包括端点的数,进行一种规律运算,并求出运算的次数,比较所有的数规律运算次数,输出最大的.
AC暴力:(hduoj的数据太弱。。。)
#include<stdio.h>#include<iostream>#include<memory.h>using namespace std;int res[1000000];int main(){ int n,m; while(scanf("%d%d",&n,&m)!=EOF){ printf("%d %d ",n,m); int temp; if(n>m){ temp=n; n=m; m=temp; } int result=0; for(int i=n;i<=m;i++){ __int64 g=i; int index=0; while(g!=1){ if(g%2==0){ g/=2; }else{ g=3*g+1; } index++; } if(result<index){ result=index; } } printf("%d\n",result+1); }}
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