(Problem 21)Amicable numbers
2013-07-24 13:11
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Let d(n) be defined as the sum of proper divisors of n (numbers less than n which divide evenly into n).
If d(a) = b and d(b) = a, where a
![](http://projecteuler.net/images/symbol_ne.gif)
b, then a and b are
an amicable pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
If d(a) = b and d(b) = a, where a
![](http://projecteuler.net/images/symbol_ne.gif)
b, then a and b are
an amicable pair and each of a and b are called amicable numbers.
For example, the proper divisors of 220 are 1, 2, 4, 5, 10, 11, 20, 22, 44, 55 and 110; therefore d(220) = 284. The proper divisors of 284 are 1, 2, 4, 71 and 142; so d(284) = 220.
Evaluate the sum of all the amicable numbers under 10000.
#include<stdio.h> #include<math.h> #include<string.h> #include<ctype.h> #include<stdlib.h> #include<stdbool.h> int FactorSum(int n) //计算n的所有小于n的因素和 { int i; int sum=1; for(i=2; i<=n/2; i++) { if(n%i==0) sum+=i; } return sum; } int main() { int t,i=2; int sum=0; while(i<10000) { t=FactorSum(i); if(t!=i && FactorSum(t)==i) sum+=i; i++; } printf("%d\n",sum); return 0; }
Answer: | 31626 |
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