PTA甲 1104. Sum of Number Segments (20)

1104. Sum of Number Segments (20)

时间限制

200 ms

内存限制

65536 kB

代码长度限制

16000 B

判题程序

Standard

作者

CAO, Peng

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:

40.1 0.2 0.3 0.4 

Sample Output:

5.00

一道很有意思的小题一开始没有考虑复杂度直接用了n方的，没想到后面两个数据过不了。后来观察规律，发现数组中数出现的次数是有规律的所以在输入的过程中就可以同时进行求和了。

ac代码

#include <stdio.h>


double a[100005];


int main()
{
int i,j,n;
double sum=0;


scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%lf",&a[i]);
double sumTmp = a[i] * (n - i) *(i + 1) ;  
        sum += sumTmp ;  
}



printf("%.2lf",sum);


return 0;


}

15分代码

#include <stdio.h>


float a[100005];


int main()
{
int i,j,n;
float sum=0,last=0,ans=0;




scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%f",&a[i]);
}


for(i=0;i<n;i++)
{
ans=0;
for(j=i;j<n;j++)
{
ans+=a[j];
sum+=ans;
}
}



printf("%.2f",sum);


return 0;


}