1104 Sum of Number Segments (20point(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 10
5
. 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
At first I thought about making a watch, but there is still the complexity of o(n^2), obviously there is a timeout
#include<bits/stdc++.h>
using namespace std;
double sequence[100010];
double sum[100010];
int main(){
int n;
cin>>n;
sum[0]=0.0;
for(int i=1;i<=n;++i){
cin>>sequence[i];
sum[i]=sum[i-1]+sequence[i];
}
double ans=0;
for(int i=1;i<=n;++i){
for(int j=i;j<=n;++j)
ans+=sum[j]-sum[i-1];
}
printf("%.2lf\n",ans);
}
Find the rules, there are basically pits in high-precision problems. After modifying the sample of this question, the double cannot be passed, and the double error becomes larger due to multiple accumulations.
#include<bits/stdc++.h>
using namespace std;
int main(){
int n;
scanf("%d",&n);
long double num,ans=0.0;
for(int i=0;i<n;++i){
scanf("%llf",&num);
ans+=num*(i+1)*(n-i);
}
printf("%.2llf\n",ans);
return 0;
}