題意:賊長,初始2,第i次操作時,可以連續加好多次i,當到達某一個完全平方數時,開更,開更後必須滿足膜(i+1)==0,求每次操作需要加多少次i
題解:根據題意 a[i] + ans[i] * i = a[i+1] * a[i+1]
ans[i] = (a[i+1] * a[i+1] - a[i]) / i
a[i] % i == 0
所以a[i+1] % i == 0
可以構造a[i] = i * i * (i-1) * (i-1)
ans[i] = (i+1)*(i+1)*i - (i-1)
int n;
int main(){
while(scanf("%d",&n)!=EOF){
for(int i=1;i<=n;i++){
if(i==1)
printf("2\n");
else
cout<<(LL)(i+1)*(i+1)*i-(i-1)<<endl;
}
}
return 0;
}