UVa10883 Supermean

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这道题开始没有注意越界问题，当 n = 5000 是取组合数非常大，而 2^n 也非常大，需要用自然对数处理一下才能处理 C(N,K)*A[k]/2^N 的求值，而且取对数的时候要分正负零各种情形考虑，提交了很多次，终于 AC 了。

package UVa10883;import java.util.Scanner;public class Main {public static double solve(double[] v, int len) {long n = len - 1;double lgc = 0, lgd = n*Math.log(2);double sum = 0, f;if ( v[0] > 0 )sum += Math.exp(Math.log(v[0]) + lgc - lgd);else if ( v[0] < 0 )sum -= Math.exp(Math.log(-v[0]) + lgc - lgd);for ( int i=1; i<=n; i++ ) {lgc = lgc + Math.log(n-i+1) - Math.log(i);if ( v[i] > 0 ) {f = Math.log(v[i]) + lgc -lgd;sum += Math.exp(f);} else if ( v[i] < 0 ) {f = Math.log(-v[i]) + lgc -lgd;sum -= Math.exp(f);}}return sum;}public static void main(String[] args) {double[] v = new double[51000];Scanner in = new Scanner(System.in);int N = in.nextInt();for ( int i=0; i<N; i++ ) {int n = in.nextInt();for ( int j=0; j<n; j++ ) {v[j] = in.nextDouble();}double r = solve(v,n);System.out.printf("Case #%d: %.3f%n", i+1, r);}}}