hdoj1032 3n+1问题

链接
http://acm.nyist.net/JudgeOnline/problem.php?pid=271
代码1 ： 每次都要调用一次函数 有些浪费时间 ！！

View Code
Problem : 1032 (The 3n + 1 problem )     Judge Status : Accepted
RunId : 10087220    Language : C    Author : sws
Code Render Status : 
Rendered By HDOJ C Code Render Version 0.01 
Beta#include<stdio.h>
int f(int x )
{
    int y = 1;
    while(x != 1 )
    {
        y ++;
    if(x%2 )
    x = 3*x+1;
    else
        x = x/2;

    }
    return y;
}
int main()
{
    int i, j, maxn, m, k, n;

    while(scanf("%d%d",&i,&j)==2){  //只有当i,j都正确读入整数时（2项）,才进行下一步
    //scanf("%d%d", &i, &j );
    maxn = 0;
    n = i;
    m = j;
    if(i>j)
    {
        k = i;
        i = j;
        j = k;
    }
    for(; i<=j; i ++ )
    if(maxn < f(i) )
        maxn = f(i);
    //n = n > m? n:m;
    printf("%d %d %d\n", n, m, maxn );
    }
    return 0;
}
#include<stdio.h>
int f(int x )
{
    int y = 1;
    while( x != 1 )
    {
        y ++;
    if( x%2 )
    x = 3*x+1;
    else
        x = x/2;

    }
    return y;
}
int main()
{
    int i, j, maxn, m, k, n;

    while(scanf("%d%d",&i,&j)==2){  //只有当i,j都正确读入整数时（2项）,才进行下一步
    //scanf("%d%d", &i, &j );
    maxn = 0;
    n = i;
    m = j;
    if( i>j)
    {
        k = i;
        i = j;
        j = k;
    }
    for(; i<=j; i ++ )
    if( maxn < f(i) )
        maxn = f(i);
    //n = n > m? n:m;
    printf( "%d %d %d\n", n, m, maxn );
    }
    return 0;
}

代码2： 再加一个函数 就解决了函数重复调用的问题 节约时间！

查看代码---运行号：724144----结果：Accepted

运行时间：2014-02-18 14:32:18 | 运行人:shengweisong

view source

print?

01.

#include<stdio.h> 
 //TL原因估计是每次都要调用函数

02.

#define
 MAX 10001

03.

int

a[MAX];

04.

int

f(

int

x
 )

05.

{

06.

int

y
 = 1;

07.

while

(
 x != 1 )

08.

{

09.

y
 ++;

10.

if

(
 x%2 )

11.

x
 = 3*x+1;

12.

else

13.

x
 = x/2;

14.

15.

}

16.

return

y;

17.

}

18.

void

CountA(){

19.

for

(

int

i
 = 1; i != MAX; ++i)

20.

a[i]
 = f(i);

21.

}

22.

int

main()

23.

{

24.

CountA();
// 如在前面加一个void 就不起作用了！！ 就在这我纠结了半天！！

25.

register

int

i,
 j, maxn, m, k, n;

26.

27.

while

(

scanf

(

"%d%d"

,&i,&j)!=EOF
 ){

//只有当i,j都正确读入整数时（2项）,才进行下一步

28.

//scanf(
 "%d%d", &i, &j );

29.

maxn
 = 0;

30.

n
 = i;

31.

m
 = j;

32.

if

(
 i>j)

33.

{

34.

k
 = i;

35.

i
 = j;

36.

j
 = k;

37.

}

38.

for

(
 ; i<=j; i ++ )

39.

if

(
 maxn < a[i] )

40.

maxn
 = a[i];

41.

//n
 = n > m? n:m;

42.

printf

(

"%d
 %d %d\n"

,
 n, m, maxn );

43.

}

44.

return

0;

45.

}