zoj 2185 || poj 1969 Count on Canton

1/1 1/2 1/3 1/4 1/5 ...

2/1 2/2 2/3 2/4

3/1 3/2 3/3

4/1 4/2

5/1

In the above diagram, the first term is 1/1, the second term is 1/2, the third term is 2/1, the fourth term is 3/1, the fifth term is 2/2, and so on.

也就是

按 1/1 1/2 2/1 3/1 2/2 1/3。。。排序。。。规律很明显。

分子都是 1  1  2  1  2  3   1  2  3  4  1.。。。

分母都是 1  2  1  3  2  1   4  3  2  1  5.。。。。

很明显了吧。不过呢 是s型排列。。。所以呢这么求出来之后需要看在斜行的奇偶性，如果在偶行，需要把分子分母颠倒下。

#include <cstdio>#include <cstdlib>#include <iostream>#include <string.h>#include <math.h>#include <queue>#include <limits.h>using namespace std;int compute(int x){int big = (int)sqrt(2*x);int i;for(i=big; i>=1; i++)if( i*(i+1)/2 >= x && i*(i-1)/2 <= x )break;int a = x - i*(i-1)/2;if( i % 2 == 1 )printf("%d/%d/n",i-a+1,a);elseprintf("%d/%d/n",a,i-a+1);}int main(){int n, ind = 1;while( ~scanf("%d",&n) ){printf("TERM %d IS ",n);compute(n);}return 0;}