思维 UVALive 3708 Graveyard
2015-04-28 15:31
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题目传送门
/* 题意:本来有n个雕塑,等间距的分布在圆周上,现在多了m个雕塑,问一共要移动多少距离; 思维题:认为一个雕塑不动,视为坐标0,其他点向最近的点移动,四舍五入判断,比例最后乘会10000即为距离; 详细解释:http://www.cnblogs.com/zywscq/p/4268556.html */ #include <cstdio> #include <iostream> #include <algorithm> #include <cmath> #include <cstring> #include <string> #include <map> #include <set> #include <queue> using namespace std; const int MAXN = 1e4 + 10; const int INF = 0x3f3f3f3f; int main(void) //UVALive 3708 Graveyard { //freopen ("UVALive_3708.in", "r", stdin); int n, m; while (scanf ("%d%d", &n, &m) == 2) { double ans = 0.0, pos; for (int i=1; i<n; ++i) { pos = (double) i / n * (n + m); ans += fabs (pos - floor (pos + 0.5)) / (n + m); } printf ("%.4lf\n", ans * 10000); } return 0; }
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