1691.Abundance

Abundance

题意:

An abundant number is a positive integer n for which Sigma(n) – 2n > 0, Where Sigma(n) is defined as the sum of all the divisors of n. And the quantity Sigma(n) – 2n is called abundance.

Given the range of n, you should find out the maximum abundance value that can be reached. For example, if the range is [10,12], then the only abundant number is 12, and the maximum abundance value is Sigma(12) – 2 * 12 = 4.

Input:

Input may contain several test cases. The first line is a positive integer, T (T<=20), the number of test cases below. Each test case contains two positive integers x, y, (1<= x <= y <= 1024), indicating the range of n.

Output

For each test case, output the maximum abundance value that can be reached in the range of n. If there is no abundant number n in the given range, simply output -1.

Sample Input

3

1 1

10 12

1 1024

Sample Output

-1

4

1208

代码实现:

// [n, m]范围内满足条件的值得最大一个
//  sigma(n) - 2*n > 0的n中， 输出最大的那个n对应的sigma(n) - 2*n
// 其中sigma(n)是n的所有因子的和

#include<iostream>
#include<cmath>
using namespace std;

int main() {
int t, n, m, num, count, total, sum;
int result, max;
int divisor[1000], fit[1000];
cin >> t;
while (t--) {
cin >> n >> m;
total = 0;
for (int i = n; i <= m; ++i) {
num = i;
count = 0;
sum = 0;
for (int j = 1; j <= num; j++) {
if (num % j == 0)
divisor[count++] = j;
}
for (int k = 0; k < count; ++k) sum += divisor[k];
result = sum - 2*num;
if (result > 0) fit[total++] = result;
}
if (total == 0) {
cout << "-1" << endl;
} else {
max = fit[0];
for (int i = 1; i < total; ++i) {
max = max > fit[i] ? max : fit[i];
}
cout << max << endl;
}
}
return 0;
}