题目地址

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

#include<iostream>
#include<string.h>
using namespace std;
int main(){     //多举例，找规律
    int n;
    double v,ans=0.0;
    scanf("%d",&n);
    for(int i=1;i<=n;i++){
        scanf("%lf",&v);
        ans+=v*i*(n-i+1);
    }
    printf("%.2lf",ans);
}
// #include <iostream>
// #include <vector>
// #include<algorithm>
// #include <cmath>
// #include<map>
// #include<cstring>
// #include<queue>
// #include<string>
// #include<set>
// #include<stack>
// using namespace std;
// typedef long long ll;
// const int maxn=110,inf=100000000;
// int n;double a[maxn],b[maxn],ans=0,t=0;
// int main(){
//     cin>>n;
//     for(int i=0;i<n;i++){
//         cin>>a[i];
//         if(i==0) b[i]=a[i];
//         else b[i]=b[i-1]+a[i];
//         t+=b[i];
//     }
//     for(int i=0;i<n;i++){
//         for(int j=i+1;j<n;j++){
//             ans+=b[j]-b[i];
//         }
//     }
//     printf("%.2lf",ans+t);
// }