ZOJ1180 Self Numbers 开始刷水题模式
2016-03-23 19:08
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In 1949 the Indian mathematician D.R. Kaprekar discovered a class of numbers called self-numbers. For any positive integer n, define d(n) to be n plus the sum of the digits of n. (The d stands for digitadition, a term coined by
Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 +
3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence
33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ...
The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators,
91 and 100. A number with no generators is a self-number. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.
Write a program to output all positive self-numbers less than or equal 1000000 in increasing order, one per line.
就是 对每个数 取模 判断 这是个水题 贴上代码
#include <iostream>
using namespace std;
int a[1000001]={0};
int d(int n)
{
int sum=n;
while(n)
{
sum+=n%10;
n/=10;
}
return sum;
}
int main()
{
int i,t;
for(int i=1;i<=1000000;i++)
{
t=d(i);
if (t <= 1000000)
{
a[t]=1;
}
}
for(int i=1;i<=1000000;i++)
{
if(!a[i])
{
cout << i<<endl;
}
}
return 0;
}
Kaprekar.) For example, d(75) = 75 + 7 + 5 = 87. Given any positive integer n as a starting point, you can construct the infinite increasing sequence of integers n, d(n), d(d(n)), d(d(d(n))), .... For example, if you start with 33, the next number is 33 +
3 + 3 = 39, the next is 39 + 3 + 9 = 51, the next is 51 + 5 + 1 = 57, and so you generate the sequence
33, 39, 51, 57, 69, 84, 96, 111, 114, 120, 123, 129, 141, ...
The number n is called a generator of d(n). In the sequence above, 33 is a generator of 39, 39 is a generator of 51, 51 is a generator of 57, and so on. Some numbers have more than one generator: for example, 101 has two generators,
91 and 100. A number with no generators is a self-number. There are thirteen self-numbers less than 100: 1, 3, 5, 7, 9, 20, 31, 42, 53, 64, 75, 86, and 97.
Write a program to output all positive self-numbers less than or equal 1000000 in increasing order, one per line.
就是 对每个数 取模 判断 这是个水题 贴上代码
#include <iostream>
using namespace std;
int a[1000001]={0};
int d(int n)
{
int sum=n;
while(n)
{
sum+=n%10;
n/=10;
}
return sum;
}
int main()
{
int i,t;
for(int i=1;i<=1000000;i++)
{
t=d(i);
if (t <= 1000000)
{
a[t]=1;
}
}
for(int i=1;i<=1000000;i++)
{
if(!a[i])
{
cout << i<<endl;
}
}
return 0;
}
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