hdu 1032 The 3n + 1 problem (打表)
來源：程序員人生 發布時間：2015-05-04 10:22:39 閱讀次數：2824次
The 3n + 1 problem
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 65536/32768 K (Java/Others)
Total Submission(s): 26353 Accepted Submission(s): 9784
題目鏈接：http://acm.hdu.edu.cn/showproblem.php?pid=1032
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.
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.
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 10
100 200
201 210
900 1000
Sample Output
1 10 20
100 200 125
201 210 89
900 1000 174
Source
UVA
在我轉的hdu題目分類
這個題目被分類為簡單題，，我仔細看了看，，數據 0 < n < 1,000,000，，， 直接做肯定超時啊，自己試了下果然1秒內算不出來，思前想后沒有辦法百度了題解，結果題解就是直接暴力做的！！ ，，我把自己暴力的代碼提交，果然ac，，數據太水了。 但是這題肯定有非暴力的方法：
analys：
暴力的時候視察到，很多數經過題目中算法描寫轉換的時候， 又回到了之前的數，例如
n=10；
10,， 5， 16， 8， 4， 2， 1；
n=11；
11,， 34， 17， 52， 26， 13， 40， 20，
10， 5， 16， 8 ，4 ，2， 1；
故總數= f【11】=f【10】+8；
所以在打表的時候可以省去很多計算；
打表法：
#include<stdio.h>
#include<algorithm>
using namespace std;
int test[2000010]={};
void init()//打表
{
for(int i=1;i<=1000000;i++)
{
long long j=i; //運算中，j可能超過int上限；
int ans=1;
while(j!=1)
{
if(j<=1000000&&test[j]!=0) //關鍵， 加上已記錄的位置的統計個數；
{
ans+=(test[j]⑴);
break;
}
else
{
if(j%2==0)
j/=2;
else
j=(j*3+1);
ans++;
}
}
test[i]=ans;
}
}
int main()
{
init();
/* for(int i=1;i<=10;i++)
printf("%d
",test[i]); */
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
printf("%d %d ",n,m);
if(n>m)
swap(n,m);
//printf("%d %d ",n,m);
int maxi=0;
for(int i=n;i<=m;i++)
{
if(test[i]>maxi)
maxi=max(maxi,test[i]);
}
printf("%d
",maxi);
}
return 0;
}
暴力代碼：
#include <stdio.h>
#include <algorithm>
using namespace std;
int main()
{
int n,m;
while(scanf("%d%d",&n,&m)!=EOF)
{
printf("%d %d ",n,m);
if(n>m)
swap(n,m);
int max=0;
for(int i=n;i<=m;i++)
{
int t=i;
int cnt=1;
while(t!=1)
{
if(t%2==0)
t/=2;
else
t=(t*3+1);
cnt++;
}
if(cnt>max)
max=cnt;
}
printf("%d
",max);
}
return 0;
}
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