NYOJ 271 The 3n + 1 problem

原题链接

这题和hdu 1032一样，令人郁闷的是在hdu ac的代码拿到nyoj就TLE了，然后果断加上打表，就过了。

附ac代码：

#include <stdio.h>
#define MAX 10001
int a[MAX];

int len( int n ){
    int i = 1;
    while( n != 1 ){
        if( n & 1 )
            n = 3 * n + 1;
        else
            n >>= 1;
        ++i;
    }
    return i;
}

void CountA(){
	for(int i = 1; i != MAX; ++i)
		a[i] = len(i);
}

int main(){
	CountA();
    int i, j, maxlen, ii, jj;
    while( scanf( "%d%d", &i, &j ) == 2 ){
         i < j ? ( ii = i, jj = j ) : ( ii = j, jj = i );
         maxlen = 0;
         while( ii <= jj ){
            int t = a[ii++];
            if( t > maxlen )
                maxlen = t;
         }
         printf( "%d %d %d\n", i, j, maxlen );
    }
    return 0;
}