UVA 11971 - Polygon(概率+几何概型)
2014-08-08 14:42
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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
;
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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