本随笔写的是第二题......
这道题方法就是搞乱....
因为n较mxa小 所以枚举达到最大上限的点 然后就乱搞 代码看看咯
#include<cstdio> #include<cstring> #include<algorithm> #define LL long long using namespace std; const int M=100005; LL read(){ LL ans=0; int 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,pos; LL v1,v2,mxa,m,sum[M],end[M],ans=-1; struct node{int id; LL v;}e[M]; bool cmp(node a,node b){return a.v<b.v;} int main() { scanf("%d",&n); mxa=read(); v1=read(); v2=read(); m=read(); for(int i=1;i<=n;i++) e[i].v=read(),e[i].id=i; sort(e+1,e+1+n,cmp); for(pos=n;pos;pos--){ sum[pos]=sum[pos+1]+(mxa-e[pos].v); if(sum[pos]>m) break; } pos++; LL now,last,use=0,rem,val; int putmn=0,putmx=0,id=0; for(int i=pos;i<=n+1;i++){ rem=m-use-sum[i]; while(id<i-1&&(val=(e[id+1].v-e[id].v)*id)<=rem) id++,rem-=val,use+=val; now=e[id].v; if(id){ now+=rem/id; now=min(now,mxa); } else now=mxa; val=v2*now+v1*(n-i+1); if(val>ans){ ans=val; last=now; putmn=id; putmx=i; } } printf("%lld ",ans); for(int i=1;i<=putmn;i++) end[e[i].id]=last; for(int i=putmn+1;i<putmx;i++) end[e[i].id]=e[i].v; for(int i=putmx;i<=n;i++) end[e[i].id]=mxa; for(int i=1;i<=n;i++) printf("%lld ",end[i]); return 0; }