UVA 138 Street Numbers

Description



A computer programmer lives in a street with houses numbered consecutively (from 1) down one side of the street. Every evening she walks her dog by leaving her house and randomly turning left or right and walking to the end of the street and back. One night
she adds up the street numbers of the houses she passes (excluding her own). The next time she walks the other way she repeats this and finds, to her astonishment, that the two sums are the same. Although this is determined in part by her house number and
in part by the number of houses in the street, she nevertheless feels that this is a desirable property for her house to have and decides that all her subsequent houses should exhibit it.

Write a program to find pairs of numbers that satisfy this condition. To start your list the first two pairs are: (house number, last number):

6         8
35        49

Input and Output

There is no input for this program. Output will consist of 10 lines each containing a pair of numbers, each printed right justified in a field of width 10 (as shown above).

题目的意思是让你找出10组n，k

满足

1+2+...+k-1 == (k+1)+...+n

已知a1，an

用等差数列求和公式Sn=n(a1+an)/2化简比较方便

(k – 1)[1 + (k – 1)] / 2 = (n – k)[(k + 1) + n] / 2,

(2n + 1)2 – 2(2k)2 = 1  
k = sqrt(n*(n+1)/2)

n循环+1

判断k不是是整数

是的话就打印出来

输出格式还有要求

宽度为10

#include <cstdio>
#include <cmath>

int main(){
int cnt = 0;
long long n = 6;
double x;
long long y;
while (cnt < 10) {
x = (double) n * (n + 1);
x /= 2;
x = sqrt(x);
y = (int)x;
if (x == y) {
printf("%10lld%10lld\n", y, n);
cnt++;
}
n++;
}

return 0;
}