UVA 11490 - Just Another Problem(数论)
2014-06-30 01:35
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11490 - Just Another Problem
题目链接题意:有S个士兵,排成一个矩阵,矩阵中可以有两个洞,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
由于S很大,所以去枚举厚度,这样只要枚举到sqrt(S)就够了,复杂度可以接受
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;
}
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