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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;
}



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