usaco2.1.2——frac1
Ordered Fractions
Consider the set of all reduced fractions between 0 and 1 inclusive with denominators less than or equal to N.
Here is the set when N = 5:
0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1
Write a program that, given an integer N between 1 and 160 inclusive, prints the fractions in order of increasing magnitude.
PROGRAM NAME: frac1
INPUT FORMAT
One line with a single integer N.SAMPLE INPUT (file frac1.in)
5
OUTPUT FORMAT
One fraction per line, sorted in order of magnitude.SAMPLE OUTPUT (file frac1.out)
0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1
描述
输入一个自然数N,请写一个程序来增序输出分母小于等于N的最简(原文:“既约”)真分数
格式
PROGRAM NAME: frac1
INPUT FORMAT:
(file frac1.in)
单独的一行 一个自然数N(1..160)
OUTPUT FORMAT:
(file frac1.out)
每个分数单独占一行,按照大小次序排列
SAMPLE INPUT
5
SAMPLE OUTPUT
0/1 1/5 1/4 1/3 2/5 1/2 3/5 2/3 3/4 4/5 1/1
USER: di zhang [codeway3]
TASK: frac1
LANG: PASCAL
Compiling...
Compile: OK
Executing...
Test 1: TEST OK [0.000 secs, 276 KB]
Test 2: TEST OK [0.000 secs, 276 KB]
Test 3: TEST OK [0.000 secs, 276 KB]
Test 4: TEST OK [0.000 secs, 276 KB]
Test 5: TEST OK [0.000 secs, 276 KB]
Test 6: TEST OK [0.000 secs, 276 KB]
Test 7: TEST OK [0.000 secs, 276 KB]
Test 8: TEST OK [0.000 secs, 276 KB]
Test 9: TEST OK [0.000 secs, 276 KB]
Test 10: TEST OK [0.000 secs, 276 KB]
Test 11: TEST OK [0.000 secs, 276 KB]
All tests OK.
YOUR PROGRAM ('frac1') WORKED FIRST TIME! That's fantastic -- and a rare thing. Please accept these special automated congratulations.
这道题的做法实在是太多了。。。
枚举+快排什么的
我这次用的是比较nb的数学方法,
叫什么也不知道。。。
如下所示
0/1 1/1
1/2
1/3 2/3
1/4 2/5 3/5 3/4
1/5 2/7 3/8 3/7 4/7 5/8 5/7 4/5
递归调用即可得解
1 2 3 4 5 6 7 8 9 10 11 12 13 14 15 16 17 18 19 20 21 22 23 24 25 26 27 28 29 | { ID: codeway3 PROG: frac1 LANG: PASCAL } program frac1; var i,j,n,m,k,l: longint ; procedure doing(a,b,c,d: longint ); var i,j: longint ; begin if (b>n) or (d>n) then exit; doing(a,b,a+c,b+d); if b+d<=n then writeln (a+c, '/' ,b+d); doing(a+c,b+d,c,d); end ; begin assign(input, 'frac1.in' ); reset(input); assign(output, 'frac1.out' ); rewrite(output); readln(n); writeln ( '0/1' ); doing( 0 , 1 , 1 , 1 ); writeln ( '1/1' ); close(input); close(output); end . |
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