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hackerrank aVeryBigSum
문제
풀이
- 큰수가 나올 수 있으니 BigInteger 사용
- 테스트 케이스 아래에 첨
package hackerrank;
import java.io.BufferedReader;
import java.io.InputStreamReader;
import java.math.BigInteger;
public class AVeryBigSum {
public static void main(String[] args) throws Exception {
BufferedReader in = new BufferedReader(new InputStreamReader(System.in));
int num = Integer.parseInt(in.readLine());
BigInteger[] bigIntegers = new BigInteger[num];
String[] str = in.readLine().split(" ");
for (int i = 0 ; i < str.length ; i++) {
bigIntegers[i] = new BigInteger(str[i]);
}
BigInteger result = aVeryBigSum(bigIntegers);
System.out.println(result);
}
public static BigInteger aVeryBigSum(BigInteger[] bigIntegers) {
BigInteger sum = BigInteger.ZERO;
for (BigInteger bigInteger : bigIntegers) {
sum = sum.add(bigInteger);
}
return sum;
}
}
public class AVeryBigSumTest {
@Test
public void aVeryBigSumTest() {
BigInteger[] bigIntegers = new BigInteger[3];
bigIntegers[0] = new BigInteger("100");
bigIntegers[1] = new BigInteger("200");
bigIntegers[2] = new BigInteger("300");
Assert.assertThat(AVeryBigSum.solve(bigIntegers), is(new BigInteger("600")));
}
}
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