HDU 4432 Sum of divisors
转载
Sum of divisors
http://acm.hdu.edu.cn/showproblem.php?pid=4432
Time Limit: 2000/1000 MS (Java/Others) Memory Limit: 32768/32768 K (Java/Others) Total Submission(s): 875 Accepted Submission(s): 342
Problem Description
mmm is learning division, she's so proud of herself that she can figure out the sum of all the divisors of numbers no larger than 100 within one day! But her teacher said "What if I ask you to give not only the sum but the square-sums of all the divisors of numbers within hexadecimal number 100?" mmm get stuck and she's asking for your help. Attention, because mmm has misunderstood teacher's words, you have to solve a problem that is a little bit different. Here's the problem, given n, you are to calculate the square sums of the digits of all the divisors of n, under the base m.
Input
Multiple test cases, each test cases is one line with two integers. n and m.(n, m would be given in 10-based) 1≤n≤109 2≤m≤16 There are less then 10 test cases.
Output
Output the answer base m.
Sample Input
Sample Output
110
112
Hint
Use A, B, C...... for 10, 11, 12...... Test case 1: divisors are 1, 2, 5, 10 which means 1, 10, 101, 1010 under base 2, the square sum of digits is 1^2+ (1^2 + 0^2) + (1^2 + 0^2 + 1^2) + .... = 6 = 110 under base 2.
Source
#include<stdio.h>
#include<math.h>
int n,m;
int bit[10000],cnt;
void change(int n,int base){
cnt=0;
while(n){
bit[cnt++]=n%base;
n/=base;
}
}
int main(){
while(scanf("%d%d",&n,&m)!=EOF){
int sum=0,i;
int t=(int)sqrt(n*1.0);
for(i=1;i<=t;i++)
if(n%i==0){
int tmp=i;
while(tmp){
sum+=(tmp%m)*(tmp%m);
tmp/=m;
}
tmp=n/i;
if(tmp==i)
continue;
while(tmp){
sum+=(tmp%m)*(tmp%m);
tmp/=m;
}
}
change(sum,m);
for(i=cnt-1;i>=0;i--)
if(bit[i]>=10)
printf("%c",bit[i]-10+'A');
else
printf("%d",bit[i]);
printf("\n");
}
return 0;
}
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