ZOJ3770Ranking System 2017-04-14 12:42 52人阅读评论(0) 收藏

Ranking System

Time Limit: 2 Seconds Memory Limit: 65536 KB

Few weeks ago, a famous software company has upgraded its instant messaging software. A ranking system was released for user groups. Each member of a group has a level placed near his
nickname. The level shows the degree of activity of a member in the group.

Each member has a score based his behaviors in the group. The level is determined by this method:

Level	Percentage	The number of members in this level
LV1	/	All members whose score is zero
LV2	/	All members who can not reach level 3 or higher but has a positive score
LV3	30%	⌊(The number of members with a positive score) * 30%⌋
LV4	20%	⌊(The number of members with a positive score) * 20%⌋
LV5	7%	⌊(The number of members with a positive score) * 7%⌋
LV6	3%	⌊(The number of members with a positive score) * 3%⌋

⌊x⌋ is the maximum integer which is less than or equal to x.
The member with the higher score will get the higher level. If two members have the same score, the earlier one who joined the group will get the higher level. If there is still a tie, the user with smaller ID will get the higher level.

Please write a program to calculate the level for each member in a group.

Input

There are multiple test cases. The first line of input is an integer T indicating the number of test cases. For each test case:

The first line contains an integer N (1 <= N <= 2000) indicating the number of members in a group.

The next N lines, each line contains three parts (separated by a space):

The ID of the i-th member A_i (0 <= A_i <= 1000000000). The ID of each member is unique.
The date of the i-th member joined the group, in the format of YYYY/MM/DD. The date will be in the range of [1900/01/01, 2014/04/06].
The score S_i (0 <= S_i <= 9999) of the i-th member.

Output

For each test case, output N lines. Each line contains a string represents the level of the i-th member.

Sample Input

1

5

123456 2011/03/11 308

123457 2011/03/12 308

333333 2012/03/18 4

555555 2014/02/11 0

278999 2011/03/18 308

Sample Output

LV3

LV2

LV2

LV1

LV2

题目的意思是给出每个等级的定义，在根据定义输出每个人的等级

思路：先统计出几个0分的，在排序之后按比例划分

#include <iostream>

#include<queue>

#include<cstdio>

#include<algorithm>

#include<cmath>

#include<set>

#include <cstring>

using namespace std;

#define LL long long

struct node

{

    int id,no,y,m,d,v,lv;

} p[100005];

bool cmp(node a,node b)

{

    if(a.v!=b.v) return a.v>b.v;

    if(a.y!=b.y) return a.y<b.y;

    if(a.m!=b.m) return a.m<b.m;

    if(a.d!=b.d) return a.d<b.d;

    return a.no<b.no;

}

bool cmp2(node a,node b)

{

    return a.id<b.id;

}

int main()

{

    int T,n;

    scanf("%d",&T);

    while(T--)

    {

        scanf("%d",&n);

        int tot=0;

        for(int i=0; i<n; i++)

        {

            p[i].id=i;

            scanf("%d %d/%d/%d %d",&p[i].no,&p[i].y,&p[i].m,&p[i].d,&p[i].v);

            if(p[i].v==0) tot++,p[i].lv=1;

        }

        sort(p,p+n,cmp);

        tot=n-tot;

        int cnt=0;

        for(int i=1; i<=tot*0.03; i++)

            p[cnt++].lv=6;

        for(int i=1; i<=tot*0.07; i++)

            p[cnt++].lv=5;

        for(int i=1; i<=tot*0.20; i++)

            p[cnt++].lv=4;

        for(int i=1; i<=tot*0.30; i++)

            p[cnt++].lv=3;

        while(p[cnt].v>0)

            p[cnt++].lv=2;

        sort(p,p+n,cmp2);

        for(int i=0; i<n; i++)

            printf("LV%d\n",p[i].lv);

    }

    return 0;

}