Problem:
Score : 200 points

Problem Statement

An integer N is a multiple of 9 if and only if the sum of the digits in the decimal representation of N is a multiple of 9.

Determine whether N is a multiple of 9.

Constraints

0 \leq N < 10^{200000}

N is an integer.

Input

Input is given from Standard Input in the following format:

N

Output

If N is a multiple of 9, print Yes; otherwise, print No.

Sample Input 1

123456789

Sample Output 1

Yes

The sum of these digits is 1+2+3+4+5+6+7+8+9=45, which is a multiple of 9, so 123456789 is a multiple of 9.

Sample Input 2

0

Sample Output 2

Yes

Sample Input 3

31415926535897932384626433832795028841971693993751058209749445923078164062862089986280

Sample Output 3

No

Solution in C (GCC 9.2.1) (judge verdict: Accepted)

#include <stdio.h>
#include <stdlib.h>
#include <string.h>

int compare(const void*a,const void *b) {
	return *(int*)a - *(int*)b;
}

int main() {
	int n, t, x, ans = 0, r[200010] = {0};
	char a[200010] = {0};
	scanf("%s", &a);
	for (int i = 0; i < 200010; i++) {
		if (a[i] == '1')r[i] = 1;
		else if (a[i] == '1')r[i] = 1;
		else if (a[i] == '2')r[i] = 2;
		else if (a[i] == '3')r[i] = 3;
		else if (a[i] == '4')r[i] = 4;
		else if (a[i] == '5')r[i] = 5;
		else if (a[i] == '6')r[i] = 6;
		else if (a[i] == '7')r[i] = 7;
		else if (a[i] == '8')r[i] = 8;
		else if (a[i] == '9')r[i] = 9;
	}
	for (int i = 0; i < 200010; i++) ans += r[i];
	if (ans % 9 == 0)printf("Yes");
	else printf("No");
	return 0;
}