Given an integer nn, Chiaki would like to find three positive integers xx, yy and zzsuch that: n=x+y+zn=x+y+z, x∣nx∣n, y∣ny∣n, z∣nz∣n and xyzxyz is maximum.

Input

There are multiple test cases. The first line of input contains an integer TT (1≤T≤1061≤T≤106), indicating the number of test cases. For each test case: The first line contains an integer nn (1≤n≤1061≤n≤106).

Output

For each test case, output an integer denoting the maximum xyzxyz. If there no such integers, output ?1?1 instead.

Sample Input

3 1 2 3

Sample Output

-1 -1 1

只有因子中有4或者有3才能被拆成 X+Y+Z=N,然后打了表驗證。
最后wa了好幾次，是因為int和int計算之后還是int就算賦值給long long .
打表代碼

#include <bits/stdc++.h> using namespace std; int main() {int T;scanf("%d", &T);while (T--){int n;for (int n = 1; n <= 100; n++){int maxt = -1;int a, b, c;for (int x = 1; x <= n; x++){for (int y = 1; y <= n - x; y++){int z = n - x - y;if (z && n % x == 0 && n % y == 0 && n % z == 0){if (maxt < x * y * z){a = x;b = y;c = z;}maxt = max(maxt, x * y * z);}}}printf("%d:%5d %d %d %d\n", n, maxt, a, b, c);if (n % 12 == 0)printf("\n");}}return 0; }

AC

#include <bits/stdc++.h> using namespace std; int main() {int T;long long n,x,y,z;long long sum;scanf("%d", &T);while (T--){scanf("%lld", &n);if ((n % 3) == 0){x = y = z = n / 3;sum = x * y * z;if (x + y + z == n)printf("%lld\n", sum);elseputs("-1");}else if ((n % 4) == 0){x = y = n / 4, z = n / 2;sum = x * y * z;if (x + y + z == n)printf("%lld\n", sum);elseputs("-1");}elseputs("-1");} }

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