uva 11490 - Just Another Problem(数学)

题目链接：uva 11490 - Just Another Problem

题目大意：有n个士兵，要排列成一个方阵，要求方阵尽量大，于是在方正的中间会空出两个正方形的区域，空出来的局域要求厚度相同，即正方形的四条边向相应方向均再有x行或者列。

解题思路：根据题意可以知道x(6x+7r)=n,x为厚度，r为正方形的边长。接着枚举x，x是n的因子。

#include <cstdio>
#include <cstring>
#include <cmath>

typedef long long ll;
const ll MOD = 100000007;

ll n;

int main () {
    while (scanf("%lld", &n) == 1 && n) {
        int cnt = 0;
        ll m = sqrt(n+0.5);
        for (ll i = 1; i <= m; i++) {
            if (n%i)
                continue;

            ll j = n / i;
            j -= 6 * i;

            if (j%7 || j <= 0)
                continue;

            ll r = (j/7) % MOD;

            ll ans = (2 * r * r) % MOD;
            if (ans == 0)
                continue;
            cnt++;
            printf("Possible Missing Soldiers = %lld\n", ans);
        }

        if (cnt == 0)
            printf("No Solution Possible\n");
        printf("\n");
    }
    return 0;
}

uva 11490 - Just Another Problem(数学),布布扣,bubuko.com