hdu4432 - Sum of divisors

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http://acm.hdu.edu.cn/showproblem.php?pid=4432

下面一种是直接模拟，一种就是进行优化了；

理解题意即可：

给你两个数：n 和m

先将n转换成m进制数，然后将各个位置上的数值平方求和，最后在按照m进制输出即可

#include "stdio.h"#include "string.h"#include "stdlib.h"#include "math.h"#include "algorithm"#include "iostream"using namespace std;int num[ 105 ] ; int main(){int n , m , i ,s;while( ~scanf( "%d%d" ,&n ,&m ) ){s = 0 ;for( i = 1 ; i <= sqrt( n ) ; i++ ){if( n % i == 0 ){int k = i ;while( k ){s += ( k % m ) * ( k % m ) ;k /= m ;}if( i * i != n ){int k = n / i ;while( k ){s += ( k % m ) * ( k % m ) ;k /= m ;}}}}num[ 0 ] = 0 ;int t = 0 ;while( s ){num[ t++ ] = s % m ;s /= m ;}t-- ;for( i = t ; i >= 0 ; i-- ){if( num[ i ] > 9 ){printf( "%c" , num[ i ] - 10 + 'A' ) ;}elseprintf( "%d" , num[ i ] ) ;}printf( "\n" );}return 0;}

#include "stdio.h"#include "stdlib.h"#include "string.h"#include "math.h"#include "iostream"#define LL long longusing namespace std;int m, buf2;LL buf[999999];LL cal(LL temp){LL buf2 = 0, res = 0;while (temp){buf2 = temp % m;temp /= m;res += (buf2 * buf2);}return res;}int main(){LL n, res;while (~scanf("%lld%d", &n, &m)){res = 0;memset(buf, 0, sizeof(buf));int i = 1;for (int j = 1; j <= sqrt(n); j++){if (!(n % j)){buf[i] = j;i += 1;res += cal(j);if (j != n / j){buf[i] = n / j;i += 1;res += cal(n / j);}}}i = 1;while (res){buf[i++] = res % m;res /= m;}for (int j = i - 1;  j >= 1; j--){if (buf[j] <= 9)printf ("%d", buf[j]);elseprintf ("%c", buf[j] + 'A' - 10);}printf ("\n");}return 0;}