UVA 11971 - Polygon(概率+几何概型)
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UVA 11971 - Polygon
题目链接
题意:给一条长为n的线段,要选k个点,分成k + 1段,问这k + 1段能组成k + 1边形的概率
思路:对于n边形而言,n - 1条边的和要大于另外那条边,然后先考虑3边和4边形的情况,根据公式在坐标系中画出来的图,总面积为x,而不满足的面积被分成几块,每块面积为x/2k,然后在观察发现一共是k + 1块,所以符合的面积为x−x∗(k+1)/2k,这样一来除以总面积就得到了概率1−(k+1)/2k
代码:
#include <cstdio>#include <cstring>const int N = 55;typedef long long ll;int t, k;ll mi[N];ll gcd(ll a, ll b) { while (b) {ll tmp = b;b = a % b;a = tmp; } return a;}int main() { mi[0] = 1; for (int i = 1; i <= 50; i++)mi[i] = mi[i - 1] * 2; scanf("%d", &t); int cas = 0; while (t--) {scanf("%*d%d", &k);printf("Case #%d: ", ++cas);if (k == 1) { printf("0/1\n"); continue;}ll zi = mi[k] - k - 1;ll mu = mi[k];ll d = gcd(zi, mu);printf("%lld/%lld\n", zi / d, mu / d); } return 0;}
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