Professor GukiZ likes programming contests. He especially likes to rate his students on the contests he prepares. Now, he has decided to prepare a new contest.

In total, n students will attend, and before the start, every one of them has some positive integer rating. Students are indexed from 1 to n. Let's denote the rating of i-th student as ai. After the contest ends, every student will end up with some positive integer position. GukiZ expects that his students will take places according to their ratings.

He thinks that each student will take place equal to . In particular, if student A has rating strictly lower then student B, A will get the strictly better position than B, and if two students have equal ratings, they will share the same position.

GukiZ would like you to reconstruct the results by following his expectations. Help him and determine the position after the end of the contest for each of his students if everything goes as expected.

The first line contains integer n (1 ≤ n ≤ 2000), number of GukiZ's students.

The second line contains n numbers a1, a2, ... an (1 ≤ ai ≤ 2000) where ai is the rating of i-th student (1 ≤ i ≤ n).

In a single line, print the position after the end of the contest for each of n students in the same order as they appear in the input.

In the first sample, students 2 and 3 are positioned first (there is no other student with higher rating), and student 1 is positioned third since there are two students with higher rating.

In the second sample, first student is the only one on the contest.

In the third sample, students 2 and 5 share the first position with highest rating, student 4 is next with third position, and students 1 and 3are the last sharing fourth position.

问你一个数排第几

#include<bits/stdc++.h>

using namespace std;

int a[];

int main(){

    int n;

    cin>>n;

    for(int i=;i<n;i++)

        cin>>a[i];

    int s=;

    for(int j=;j<n;j++)

        if(a[j]>a[])s++;

    printf("%d",s);

    for(int i=;i<n;i++){

        s=;

        for(int j=;j<n;j++)

        if(a[j]>a[i])s++;

    printf(" %d",s);}

    return ;

}

直接这样枚举也能过，所以，sort的也能过吧

#include<bits/stdc++.h>

using namespace std;

int a[],b[];

int main(){

    int n;

    cin>>n;

    for(int i=;i<n;i++)

        cin>>a[i];

    copy(a,a+n,b);

    sort(b,b+n);

    cout<<n-(upper_bound(b,b+n,a[])-b)+;

    for(int i=;i<n;i++)

        cout<<" "<<n-(upper_bound(b,b+n,a[i])-b)+;

    return ;

}

完美，直接upper_bound美滋滋

aaa
a
b

aaa

pozdravstaklenidodiri
niste
dobri

nisteaadddiiklooprrvz

abbbaaccca
ab
aca

ababacabcc

给定A,B,C串，其中A串(可能)包含B或C串，现在可以任意变换A串的位置，使A串中包含B串，C串的总个数最大。

所以贪心下啊，统计字母个数

#include <bits/stdc++.h>

using namespace std;

char a[],b[],c[];

int A[],B[],C[],S[];

int main() {

    cin>>a>>b>>c;

    int len=strlen(a);

    for(int i=; i<len; i++)

        A[a[i]-'a']++;

    len=strlen(b);

    for(int i=; i<len; i++)

        B[b[i]-'a']++;

    len=strlen(c);

    for(int i=; i<len; i++)

        C[c[i]-'a']++;

    int ans,x,y;

    ans=x=y=;

    len=strlen(a)/strlen(b);

    for(int i=; i<=len; i++) {

        int ok=;

        memcpy(S,A,sizeof(A));

        for(int j=; j<; j++) {

            if(B[j]*i>S[j]) {

                ok=;

                break;

            }

            S[j]-=B[j]*i;

        }

        if(!ok)

            continue;

        int t=;

        for(int j=; j<; j++) {

            if(C[j]==)

                continue;

            t=min(t,S[j]/C[j]);

        }

        if(t+i>ans) {

            ans=t+i;

            x=i;

            y=t;

        }

    }

    for(int i=; i<x; i++)

        cout<<b;

    for(int i=; i<y; i++)

        cout<<c;

    for(int i=; i<; i++) {

        for(int j=; j<A[i]-x*B[i]-y*C[i]; j++)

            cout<<char('a'+i);

    }

    cout<<endl;

    return ;

}