1104 Sum of Number Segments (20 point(s))
Given a sequence of positive numbers, a segment is defined to be a consecutive subsequence. For example, given the sequence { 0.1, 0.2, 0.3, 0.4 }, we have 10 segments: (0.1) (0.1, 0.2) (0.1, 0.2, 0.3) (0.1, 0.2, 0.3, 0.4) (0.2) (0.2, 0.3) (0.2, 0.3, 0.4) (0.3) (0.3, 0.4) and (0.4).
Now given a sequence, you are supposed to find the sum of all the numbers in all the segments. For the previous example, the sum of all the 10 segments is 0.1 + 0.3 + 0.6 + 1.0 + 0.2 + 0.5 + 0.9 + 0.3 + 0.7 + 0.4 = 5.0.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N, the size of the sequence which is no more than 105. The next line contains N positive numbers in the sequence, each no more than 1.0, separated by a space.
Output Specification:
For each test case, print in one line the sum of all the numbers in all the segments, accurate up to 2 decimal places.
Sample Input:
4
0.1 0.2 0.3 0.4
Sample Output:
5.00
经验总结:
这一题,就是纯粹的找规律了。。。不过找不到规律也不要紧,可以直接暴力枚举,至少可以通过前两个测试点,得到15分,不过还要注意,找到规律,写公式的时候 v*i*(n+1-i) 这里千万别写成 v*(i* (n+1-i)) ,否则也只能通过前两个测试点,主要原因,猜测是这一题可能依赖浮点运算的精度,如果先计算后面两个,可能导致最后计算结果有些许误差,和判题机所给答案不同而判错,注意一下就好~
AC代码
#include <cstdio>
int n;
int main()
{
scanf("%d",&n);
double ans=0,t;
for(int i=1;i<=n;++i)
{
scanf("%lf",&t);
ans+=t*i*(n+1-i);
}
printf("%.2f\n",ans);
return 0;
}