显然发现可以从右往左依次确定。
考虑求出前缀和,每次查询当前位置上的值,然后删掉当前数,动态维护一下前缀和就好了
对于查询值可以树状数组+二分\(O(n\log^2n)\),也可以线段树\(O(n\log n)\)
貌似树状数组+二分比线段树跑的要快
代码:
#include#include #include #include using namespace std;#define rg registervoid read(long long &x){ char ch;bool ok; for(ok=0,ch=getchar();!isdigit(ch);ch=getchar())if(ch=='-')ok=1; for(x=0;isdigit(ch);x=x*10+ch-'0',ch=getchar());if(ok)x=-x;}const int maxn=2e5+10;int n,m,ans[maxn];long long f[maxn],a[maxn];#define lowbit(i) (i&(-i))void add(int x,int y){for(rg int i=x;i<=n;i+=lowbit(i))f[i]+=y;}long long get(int x){long long ans=0;for(rg int i=x;i;i-=lowbit(i))ans+=f[i];return ans;}bool check(int x,long long y){ long long now=get(x); return now<=y;}int main(){ scanf("%d",&n); for(rg int i=1;i<=n;i++)read(a[i]),add(i,i); for(rg int i=n;i;i--){ int l=1,r=n; while(l<=r){ int mid=(l+r)>>1; if(check(mid,a[i]))l=mid+1; else r=mid-1; } ans[i]=l;add(l,-l); } for(rg int i=1;i<=n;i++)printf("%d ",ans[i]);}