假设数组为a[]，直接利用动归来求解，考虑到可能存在负数的情况，我们用Max来表示以a结尾的最大连续子串的乘积值，用Min表示以a结尾的最小的子串的乘积值，那么状态转移方程为：Max=max{a, Max[i-1]*a, Min[i-1]*a};Min=min{a, Max[i-1]*a, Min[i-1]*a};初始状态为Max[1]=Min[1]=a[1]

 1 package longest_multiple; 2  3 import java.util.Scanner; 4  5  6 public class Main { 7      8     final static int N = 100001; 9     static double a[] = new double[N];10     static double Min[] = new double[N];11     static double Max[] = new double[N];12 13     static double max(double a, double b) {14         return (a > b) ? a : b;15     }16 17     static double min(double a, double b) {18         return (a < b) ? a : b;19     }20 21     static double longest_multile(double a[], int n) {22 23         Max[0] = Min[0] = a[0];24         double max_value = http://www.mamicode.com/a[0];25         for (int i = 1; i < n; i++) {26 27             Max[i] = max(max(a[i], Max[i - 1] * a[i]), Min[i - 1] * a[i]);28             Min[i] = min(min(a[i], Max[i - 1] * a[i]), Min[i - 1] * a[i]);29             if (max_value < Max[i]) {30                 max_value =http://www.mamicode.com/ Max[i];31             }32         }33         return max_value;34     }35 36     public static void main(String[] args) {37         Scanner sc = new Scanner(System.in);38         while (sc.hasNext()) {39 40             int n = sc.nextInt();41 42             for (int i = 0; i < n; i++) {43                 a[i] = sc.nextDouble();44             }45             double res = longest_multile(a, n);46             if (res < 0) {47                 System.out.println("-1");48             } else {49                 if (res - (int) res <= 1e-5) {50                     System.out.println((int) res);51                 } else {52 53                     System.out.printf("%.2lf\n", res);54                 }55             }56         }57     }58 59 }

 1 #include <stdio.h> 2   3 #define LEN 100000 4   5 int N; 6 double data[LEN]; 7 double max[LEN]; 8 double min[LEN]; 9  10 double Max(double a, double b){11     return (a > b) ? a : b;12 }13  14 double Min(double a, double b){15     return (a < b) ? a : b;16 }17  18 double MMS(){19     int i;20     double ans;21     max[0] = min[0] = data[0];22     ans = data[0];23     for (i = 1; i < N; ++i){24         max[i] = Max(Max(data[i], max[i-1]*data[i]), min[i-1]*data[i]);25         min[i] = Min(Min(data[i], max[i-1]*data[i]), min[i-1]*data[i]);26         ans = Max(ans, max[i]);27     }28     return ans;29 }30  31 int main(void){32     int i;33     double ans;34     while (scanf("%d", &N) != EOF){35         for (i = 0; i < N; ++i)36             scanf("%lf", &data[i]);37         ans = MMS();38         if (ans < 0)39             printf("-1\n");40         else if (ans - (int)ans <= 1e-5)41             printf("%lld\n", (int)ans);42         else43             printf("%.2lf\n", ans);44     }45  46     return 0;47 }