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
code
遍歷
time out
// time out error
#include <iostream>
using namespace std;
int main()
{
int n;
float num[100000];
int srt;
int ed;
float sum = 0;
scanf("%d\n",&n);
for(int i=0;i<n;i++)
{
scanf("%f",&num[i]);
}
for(srt=0;srt<n;srt++)
{
for(ed = srt;ed<n;ed++)
{
for(int j=srt;j<=ed;j++)
{
sum = sum + num[j];
}
}
}
printf("%.2f",sum);
}
正確版
#include <iostream>
using namespace std;
int main()
{
int n;
cin >> n;
double num[100001];
double sum = 0;
for(int i=1;i<=n;i++)
{
cin >> num[i];
sum = sum + num[i]*i*(n-i+1);
}
printf("%.2f",sum);
}
note
數(shù)學(xué)問(wèn)題
