HDU 1032.The 3n + 1 problem【注意细节】【预计数据不强】【8月21】
The 3n + 1 problem
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.
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.
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).
1 10
100 200
201 210
900 1000
1 10 20
100 200 125
201 210 89
900 1000 174
预计数据弱,不然直接做应该会超时(本来想写优化来着,没想到这题数据这么弱······)。
代码例如以下:
#include<cstdio>
int main(){
int n,m;
while(scanf("%d%d",&n,&m)==2){
int maxn=n>m?n:m;//没说n和m的大小,坑啊! int minx=n>m?m:n;
int maxx=0;
for(int i=minx;i<=maxn;i++){//总感觉这样会超时。看来数据不强啊
int x=i,sum=1;
while(x!=1){
if(x%2==0) x/=2;
else x=3*x+1;
sum++;
}
if(sum>maxx) maxx=sum;
}
printf("%d %d %d\n",n,m,maxx);
}
return 0;
}
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