UVA 11490 - Just Another Problem(数论)

11490 - Just Another Problem

option=com_onlinejudge&Itemid=8&page=show_problem&category=474&problem=2485&mosmsg=Submission+received+with+ID+13809788" target="_blank" style="">题目链接

题意：有S个士兵。排成一个矩阵，矩阵中能够有两个洞，要求两个洞上下左右厚度一样，问能缺少士兵的情况数。

思路：推推公式，设厚度为a, 正方形为i, 那么(3 a + 2 i) (2
a + i) = S + 2 i i;

化简一下得到6 i i + 7 a i = S

因为S非常大，所以去枚举厚度。这样仅仅要枚举到sqrt(S)就够了，复杂度能够接受

代码：

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

const long long MOD =100000007;
long long n;

int main() {
	while (~scanf("%lld", &n) && n) {
		int flag = 1;
		for (long long i = 1; i * i * 6 < n; i++) {
			long long tmp = n - i * i * 6;
			if (tmp % (7 * i) == 0) {
				long long ans = tmp / (7 * i) % MOD;
				printf("Possible Missing Soldiers = %lld\n", ans * ans * 2 % MOD);
				flag = 0;
   			}
  		}
  		if (flag) printf("No Solution Possible\n");
  		printf("\n");
 	}
	return 0;
}