算法思考: poj 1969 Count on Canton

                                  A - Count on Canton

Time Limit:1000MS     Memory Limit:30000KB     64bit IO Format:%I64d & %I64u

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Description

One of the famous proofs of modern mathematics is Georg Cantor's demonstration that the set of rational numbers is enumerable. The proof works by using an explicit enumeration of rational numbers as shown in the diagram below.

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.

Input

The input list contains a single number per line and will be terminated by endof-file.

Output

You are to write a program that will read a list of numbers in the range from 1 to 10^7 and will print for each number the corresponding term in Cantor's enumeration as given below.

Sample Input

3
14
7

Sample Output

TERM 3 IS 2/1
TERM 14 IS 2/4
TERM 7 IS 1/4



代码：
    #include <stdio.h>
#include <string.h>

int main()
{
    int n;
    int i, j;
    int dd, ff;
    int d, c;

    while(scanf("%d", &n)!=EOF )
    {
        d = 1;
        while(( (1+d)*d)/2 < n)
        {
            d++;
        }
        c = d-1;
        if(d%2==1)//奇数列
        {
            dd = d-( n-(1+c)*c/2 )+1 ;
            ff = n-(1+c)*c/2;
        }
        else if(d%2==0)//偶数列
        {
            dd = n-(1+c)*c/2 ;
            ff = d-( n-(1+c)*c/2 ) +1;
        }
        printf("TERM %d IS %d/%d
", n, dd, ff );
    }
    return 0;
}