Submission #73615370

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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <inttypes.h>
int cmp(const void* x, const void* y) {
  int a = *(const int*)x, b = *(const int*)y;
  return (a > b) - (a < b);
}
int N, M;
int A[212345];
int meow[12][212345];
int main(void) {
  int i, j;
  int64_t ans = 0;
  if (scanf("%d%d", &N, &M) != 2) return 1;
  for (i = 0; i < N; i++) {
    int cur;
 
 
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#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <inttypes.h>

int cmp(const void* x, const void* y) {
	int a = *(const int*)x, b = *(const int*)y;
	return (a > b) - (a < b);
}

int N, M;
int A[212345];

int meow[12][212345];

int main(void) {
	int i, j;
	int64_t ans = 0;
	if (scanf("%d%d", &N, &M) != 2) return 1;
	for (i = 0; i < N; i++) {
		int cur;
		if (scanf("%d", &A[i]) != 1) return 1;
		for (cur = A[i] % M, j = 0; j < 12; j++) {
			meow[j][i] = cur;
			cur = (int)((long long)cur * 10 % M);
		}
	}
	for (i = 0; i < 12; i++) {
		qsort(meow[i], N, sizeof(*meow[i]), cmp);
	}
	for (i = 0; i < N; i++) {
		char str[16];
		int len, target;
		int l, ge, le, g;
		snprintf(str, sizeof(str), "%d", A[i]);
		len = (int)strlen(str);
		target = (M - A[i] % M) % M;
		l = -1;
		ge = N;
		while (l + 1 < ge) {
			int m = l + (ge - l) / 2;
			if (meow[len][m] >= target) ge = m; else l = m;
		}
		le = -1;
		g = N;
		while (le + 1 < g) {
			int m = le + (g - le) / 2;
			if (meow[len][m] > target) g = m; else le = m;
		}
		ans += le - ge + 1;
	}
	printf("%" PRId64 "\n", ans);
	return 0;
}

/*

1. 10^n 倍してMで割った余りを求める
2. f(x, y) の y として各値を使ったとき、残り何なら M の倍数か求める
3. M で割った余りがそれになっているやつの個数を二分探索で求める

*/

Submission Info

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