考虑a排在b前边的条件
at*bv<bt*av
即at/av<bt/bv
排序即可
#include<iostream>#include<cstdio>#include<cstdlib>#include<cstring>#include<ctime>#include<cmath>#include<algorithm>#include<iomanip>#include<vector>#include<map>#include<set>#include<bitset>#include<queue>#include<stack>using namespace std;#define MAXN 100010#define MAXM 1010#define INF 1000000000#define MOD 1000000007#define eps 1e-8#define ll long longstruct zcc{int x;int y;double k;friend bool operator <(zcc x,zcc y){return x.k<y.k;}};int n;zcc a[MAXN];ll ans;ll now;int main(){int i;scanf("%d",&n);for(i=1;i<=n;i++){scanf("%d",&a[i].x);}for(i=1;i<=n;i++){scanf("%d",&a[i].y);a[i].k=1.0*a[i].x/a[i].y;}sort(a+1,a+n+1);for(i=1;i<=n;i++){now+=a[i].x;ans+=now*a[i].y;}printf("%lld\n",ans);return 0;}/**/
