1104. Sum of Number Segments (20)

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) (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

num 输入队列有多少个数
依次输入num个in，都是不大于1.0的小数，
要求输出这个队列非空子串的和，保留两位小数，有没有换行符结束都可以通过。
注意：如果 sum += (num - i + 1)*i*in;就有两个错误，详见后面。double型in要在前面，可能是数据大溢出了吧
串的各部分术语http://xujiayu317.blog.163.com/blog/static/25475209201601895222262/
评测结果

时间	结果	得分	题目	语言	用时(ms)	内存(kB)	用户
6月05日 11:43	答案正确	20	1104	C++ (g++ 4.7.2)	119	628	datrilla

测试点

测试点	结果	用时(ms)	内存(kB)	得分/满分
0	答案正确	3	512	12/12
1	答案正确	4	512	3/3
2	答案正确	119	604	3/3
3	答案正确	63	628	2/2

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
double sum=0, in;
int num;
cin >> num;
for (int i = num; i > 0; i--)
{
cin >> in;
sum += in*(num - i + 1)*i;
}
cout << setiosflags(ios::fixed) << setprecision(2)
<< sum ;
system("pause");
return 0;
}

评测结果

时间	结果	得分	题目	语言	用时(ms)	内存(kB)	用户
6月05日 11:42	部分正确	15	1104	C++ (g++ 4.7.2)	96	624	datrilla

测试点

测试点	结果	用时(ms)	内存(kB)	得分/满分
0	答案正确	3	512	12/12
1	答案正确	3	512	3/3
2	答案错误	92	512	0/3
3	答案错误	96	624	0/2

#include<iostream>
#include<iomanip>
using namespace std;
int main()
{
double sum=0, in;
int num;
cin >> num;
for (int i = num; i > 0; i--)
{
cin >> in;
sum += (num - i+1)*i*in;
}
cout << setiosflags(ios::fixed) << setprecision(2)
<< sum ;
system("pause");
return 0;
}