哈理工校园编程练习赛杭电 acm 4432 G.Sum of divisors
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Sum of divisors
TimeLimit: 2000/1000 MS(Java/Others) Memory Limit:32768/32768 K (Java/Others)
ProblemDescription
mmmis learning division, she's so proud of herself that she canfigure out the sumof all the divisors of numbers no larger than 100 within oneday!
But her teacher said "What if I ask you to give not only the sum butthesquare-sums of all the divisors of numbers within hexadecimal number100?"mmm get stuck and she's asking for your help.
Attention, because mmm has misunderstood teacher's words, you have to solveaproblem that is a little bit different.
Here's the problem, given n, you are to calculate the square sums of thedigitsof all the divisors of n, under the base m.
Input
Multipletest cases, each test cases is one line with twointegers.
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
Outputthe answer base m.
SampleInput
10 2
30 5
SampleOutput
110
112
Hint
Use A, B, C...... for 10, 11, 12......
Test case 1: divisors are 1, 2, 5,10which 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.
这个题压根就没时间做 做也做不出来啊 可笨蛋了现在的我 丫的
#include<iostream>
#include<math.h>
usingnamespace std;
intn,m;
intbit[10000],cnt;
voidchange(int n,int base)
{
cnt=0;
if(n)
{
change(n/base,base);
printf("%c",n%base>9?n%base+'A'-10:n%base+'0');
}
}
intmain()
{
while(cin>>n>>m)
{
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);
cout<<endl;
}
return 0;
}
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