公式

有n个数，分别为a $1_{}$1​, a $2_{}$2​, a $3_{}$3​, a $4_{}$4​, …, a $n_{}$n​
设 $ans(x,y)=\sum _{i=x}^{y-1}\sum _{j=i+1}^y{a}_i\ast a_j$ans(x,y)=i=x∑y−1​j=i+1∑y​ai​∗aj​
$sum\left(x\right)=\sum _{i=1}^x{a}_i$sum(x)=i=1∑x​ai​
$cost\left(y\right)=ans(1,y)=\sum _{i=1}^{y-1}\sum _{j=i+1}^y{a}_i\ast a_j$cost(y)=ans(1,y)=i=1∑y−1​j=i+1∑y​ai​∗aj​
$结论:ans(x,y)=cost\left[y\right]-cost[x-1]-sum[x-1]\ast \left(sum\right[y]-sum[x-1\left]\right))$结论:ans(x,y)=cost[y]−cost[x−1]−sum[x−1]∗(sum[y]−sum[x−1]))

for(int i = 1; i <= n; i++){
	scanf("%lld", &val[i]);
   	sum[i] = sum[i - 1] + val[i];
  }
for(int i = 1; i <= n; i++){
   	for(int j = i + 1; j <= n; j++){
   		 cost[j] += val[i] * val[j];
  	 }
}
for(int i = 1; i <= n; i++){
        cost[i] += cost[i - 1];
 }