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Our Chef is catering for a big corporate office party and is busy preparing different mouth watering dishes. The host has insisted that he serves his delicious cupcakes for dessert.
On the day of the party, the Chef was over-seeing all the food arrangements as well, ensuring that every item was in its designated position. The host was satisfied with everything except the cupcakes. He noticed they were arranged neatly in the shape of a rectangle. He asks the Chef to make it as square-like as possible.
The Chef is in no mood to waste his cupcakes by transforming it into a perfect square arrangement. Instead, to fool the host, he asks you to arrange the N cupcakes as a rectangle so that the differencebetween the length and the width is minimized.
The first line of the input file contains an integer T, the number of test cases. Each of the following T lines contains a single integer N denoting the number of cupcakes.
Output T lines, each indicating the minimum possible difference between the length and the width in a rectangular arrangement of the cupcakes.
1 ≤ T ≤ 100
1 ≤ N ≤ 108
Input:4201384Output:11220
一题简单的因子分解题目。
思路有两个：
1 从根号2開始分解，第一个能够分解的两个因子就是答案
2 利用素数，找出全部的因子，然后求解
这里使用第二中方法。这个比較有挑战性。
#include#include #include using namespace std;class ArrangingCupcakes{ const static int MAX_V = 10001; vector sieve; void genSieve() { bool A[MAX_V] = {false}; for (int i = 2; i < MAX_V; i++) { if (!A[i]) { for (int j = (i<<1); j < MAX_V; j+=i) { A[j] = true; } sieve.push_back(i); } } }public: ArrangingCupcakes() : sieve() { genSieve(); int T = 0, N = 0; scanf("%d", &T); while (T--) { scanf("%d", &N);//写成sanf("%d", N)就会程序崩溃 vector divs(1, 1); int n = N; for (int i = 0; i < (int)sieve.size() && n > 1; i++) { int sq = sieve[i]; if (sq * sq > n) sq = n; if (n % sq == 0) { int degree = 0; for ( ; n % sq == 0; degree++) n /= sq; for (int j = int(divs.size()) - 1; j >= 0 ; j--) { int d = divs[j]; for (int k = 0; k < degree; k++) { d *= sq; divs.push_back(d); } } }//if (n % sq == 0) }//求因子分解 int ans = N - 1; for (int i = 0; i < (int)divs.size(); i++) { ans = min(ans, abs(N/divs[i] - divs[i])); } printf("%d\n", ans); }//while (T--) }};
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