It is a common knowledge the importance in practicing a physical activity, mainly when you take part in ICPC competitions. Keeping this in mind, Giovana Delfino invited her friends from MaratonIME to go rowing.
The initial total ability of the boat is 0. Despite the equal enthusiasm of all members, each friend i has a physical ability h(i). Dear teacher Gabi knows that whenever two friends x and y are together in the boat, the boat's total ability is increased by h(x)·h(y).
Giovana invited n friends, but rowing is not always possible when you have programming assignments and exams of the great professor Arnaldo Mandel ahead, so sometimes not everyone shows up. However, the teacher Gabi is very optimistic and hopes that, for the final semester competition, all n friends will attend in order to increase MaratonIME odds. Help Gabi find the total ability of the boat in this possible golden moment, assuming all friends show up.
Input
The first line contains one integer n (1 ≤ n ≤ 105) - the number of friends invited to row.
The second line contains n integers h1, h2, ..., hn (1 ≤ hi ≤ 3·104) - the abilities of each member.
Output
Print one integer - the boat's total ability in the golden moment.
Examples
Input
2 5 10
Output
50
Input
4 1 3 7 11
Output
152
Note
In the first example, in the golden moment there will be two members on the boat, then the ability there will be 5 × 10 = 50.
题目大意及思路：给你n个数，让你求第i个数和后面的数的乘积。
反思：直接用的暴力，超时了，最后也没想出来用什么办法。结束后，想了想，可以用一个数组，来存储后面的数的总和，最后的复杂度就低了O（n）
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
struct node
{
int x;
long long int s;
} st[100100];
int main()
{
long long int sum;
int n, i;
sum = 0;
scanf("%d", &n);
scanf("%d", &st[0].x);
st[0].s = st[0].x;
for(i = 1; i < n; i++)
{
scanf("%d", &st[i].x);
st[i].s = st[i].x + st[i-1].s;
}
for(i = 0; i < n; i++)
{
sum = sum + st[i].x * (st[n-1].s - st[i].s);
}
printf("%lld\n", sum);
return 0;
}