算法思考: poj 1969 Count on Canton
2024-09-02 16:14:34
A - Count on Canton
Time Limit:1000MS Memory Limit:30000KB 64bit IO Format:%I64d & %I64u
Submit Status
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.
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", n, dd, ff );
}
return 0;
}
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