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2f2e9442
编写于
9月 30, 2020
作者:
S
Sombit Bose
提交者:
GitHub
9月 30, 2020
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差异文件
Create Problem12.java
Added solution for problem 12 of Project Euler
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ProjectEuler/Problem12.java
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/*
The sequence of triangle numbers is generated by adding the natural numbers. So the 7th triangle number would be 1 + 2 + 3 + 4 + 5 + 6 + 7 = 28.
The first ten terms would be:
1, 3, 6, 10, 15, 21, 28, 36, 45, 55, ...
Let us list the factors of the first seven triangle numbers:
1: 1
3: 1,3
6: 1,2,3,6
10: 1,2,5,10
15: 1,3,5,15
21: 1,3,7,21
28: 1,2,4,7,14,28
We can see that 28 is the first triangle number to have over five divisors.
What is the value of the first triangle number to have over five hundred divisors?
*/
public
class
Problem_12_Highly_Divisible_Triangular_Number
{
/* returns the nth triangle number; that is, the sum of all the natural numbers less than, or equal to, n */
public
static
int
triangleNumber
(
int
n
)
{
int
sum
=
0
;
for
(
int
i
=
0
;
i
<=
n
;
i
++)
sum
+=
i
;
return
sum
;
}
public
static
void
main
(
String
[]
args
)
{
long
start
=
System
.
currentTimeMillis
();
// start the stopwatch
int
j
=
0
;
// j represents the jth triangle number
int
n
=
0
;
// n represents the triangle number corresponding to j
int
numberOfDivisors
=
0
;
// number of divisors for triangle number n
while
(
numberOfDivisors
<=
500
)
{
// resets numberOfDivisors because it's now checking a new triangle number
// and also sets n to be the next triangle number
numberOfDivisors
=
0
;
j
++;
n
=
triangleNumber
(
j
);
// for every number from 1 to the square root of this triangle number,
// count the number of divisors
for
(
int
i
=
1
;
i
<=
Math
.
sqrt
(
n
);
i
++)
if
(
n
%
i
==
0
)
numberOfDivisors
++;
// 1 to the square root of the number holds exactly half of the divisors
// so multiply it by 2 to include the other corresponding half
numberOfDivisors
*=
2
;
}
long
finish
=
System
.
currentTimeMillis
();
// stop the stopwatch
System
.
out
.
println
(
n
);
System
.
out
.
println
(
"Time taken: "
+
(
finish
-
start
)
+
" milliseconds"
);
}
}
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