这道题很容易想到优先把两堆重量最小的合并比较优 然后乱搞一下就可以啦
#include<cstdio> #include<cstring> #include<algorithm> #include<queue> using namespace std; int read(){ int ans=0,f=1,c=getchar(); while(c<'0'||c>'9'){if(c=='-') f=-1; c=getchar();} while(c>='0'&&c<='9'){ans=ans*10+(c-'0'); c=getchar();} return ans*f; } int n; struct node{ int w; bool operator<(const node&x)const{return x.w<w;} }; int tot; priority_queue<node>q; int main() { int ans; n=read(); for(int i=1;i<=n;i++) q.push((node){read()}); while(1){ node x=q.top(); ans=x.w; q.pop(); if(q.empty()){printf("%d ",tot); return 0;} node y=q.top(); ans+=y.w; q.pop(); q.push((node){ans}); tot+=ans; } return 0; }