The ranklist of PAT is generated from the status list, which shows the scores of the submissions. This time you are supposed to generate the ranklist for PAT.

Input Specification:

Each input file contains one test case. For each case, the first line contains 3 positive integers, $N$ (≤10​4​​), the total number of users, $K$

($\le 5$), the total number of problems, and $M$ (≤10​5​​), the total number of submissions. It is then assumed that the user id's are 5-digit numbers from 00001 to $N$, and the problem id's are from 1 to $K$. The next line contains $K$ positive integers p[i] (i=1, ..., $K$), where p[i] corresponds to the full mark of the i-th problem. Then $M$ lines follow, each gives the information of a submission in the following format:

user_id problem_id partial_score_obtained

where partial_score_obtained is either $-1$ if the submission cannot even pass the compiler, or is an integer in the range [0, p[problem_id]]. All the numbers in a line are separated by a space.

Output Specification:

For each test case, you are supposed to output the ranklist in the following format:

rank user_id total_score s[1] ... s[K]

where rank is calculated according to the total_score, and all the users with the same total_score obtain the same rank; and s[i] is the partial score obtained for the i-th problem. If a user has never submitted a solution for a problem, then "-" must be printed at the corresponding position. If a user has submitted several solutions to solve one problem, then the highest score will be counted.

The ranklist must be printed in non-decreasing order of the ranks. For those who have the same rank, users must be sorted in nonincreasing order according to the number of perfectly solved problems. And if there is still a tie, then they must be printed in increasing order of their id's. For those who has never submitted any solution that can pass the compiler, or has never submitted any solution, they must NOT be shown on the ranklist. It is guaranteed that at least one user can be shown on the ranklist.

Sample Input:

7 4 20
20 25 25 30
00002 2 12
00007 4 17
00005 1 19
00007 2 25
00005 1 20
00002 2 2
00005 1 15
00001 1 18
00004 3 25
00002 2 25
00005 3 22
00006 4 -1
00001 2 18
00002 1 20
00004 1 15
00002 4 18
00001 3 4
00001 4 2
00005 2 -1
00004 2 0

Sample Output:

1 00002 63 20 25 - 18
2 00005 42 20 0 22 -
2 00007 42 - 25 - 17
2 00001 42 18 18 4 2
5 00004 40 15 0 25 -
code：

#include <stdio.h>
#include <string.h>
#include <stdlib.h>
int n,k,m;//n使用者數量，k題目個數，m提交資訊個數
int fullmark[6];//每個題目的滿分分數

struct user{
    int id;//使用者id
    int sflag;//標記是否提交併且通過了編譯，是否輸出
    int total_score;//總分
    int fullnumber;//滿分個數
    int p[6];//每個題的分數
    int pflag[6];//標記每個題目，是輸出-還是數字
}stu[10005];

void init(){
    int i,j;
    for(i = 1; i <= n; i++){
        stu[i].id = i;
        stu[i].sflag = 0;
        stu[i].fullnumber = 0;
        stu[i].total_score = 0;
        for(j = 1; j <= k; j++){
            stu[i].p[j] = 0;
            stu[i].pflag[j] = 0;
        }
    }
}

int cmp(const void *a,const void *b){
    struct user* c = (struct user*)a;
    struct user* d = (struct user*)b;
    if(c->total_score == d->total_score){
        if(c->fullnumber == d->fullnumber){
            return c->id > d->id;
        }
        return c->fullnumber < d->fullnumber;
    }
    return c->total_score < d->total_score;
}

int main(){
    int i,j;
    scanf("%d%d%d",&n,&k,&m);
    init();
    for(i = 1; i <= k; i++){
        scanf("%d",&fullmark[i]);
    }//輸入每個題的總分
    for(i = 1; i <= m; i++){
        int id,pid,score;
        scanf("%d%d%d",&id,&pid,&score);
        if(score > -1){
            stu[id].sflag = 1;//首先看分數，如果分數大於-1，說明這個使用者已經編譯通過了一次，那麼輸出rank時要輸出他
            if(score > stu[id].p[pid]){//如果新提交的分數大於原來分數，更新這道題的分數
                stu[id].p[pid] = score;
            }
        }
        stu[id].pflag[pid] = 1;//這道題提交過了應該輸出分數，不是'-'
    }
    for(i = 1; i <= n; i++){//算總分和滿分數
        if(!stu[i].sflag){
            stu[i].total_score = -2;
            continue;
        }
        for(j = 1; j <= k; j++){
            stu[i].total_score += stu[i].p[j];//算總分每個加起來即可
            if(stu[i].p[j] == fullmark[j]){//如果有滿分的，滿分數加一
                stu[i].fullnumber++;
            }
        }
    }
    //排序
    qsort(stu+1,n,sizeof(stu[1]),cmp);
    //輸出
    int ranks = 1;
    int num = 0;
    for(i = 1; i <= n; i++){
        num++;
        if(!stu[i].sflag)break;
        if(i == 1){
            printf("%d",ranks);
        }
        else{
            if(stu[i].total_score == stu[i-1].total_score){
                printf("%d",ranks);
            }
            else{
                ranks = num;
                printf("%d",ranks);
            }
        }//輸出rank
        printf(" %05d %d",stu[i].id,stu[i].total_score);
        for(j = 1; j <= k; j++){
            if(!stu[i].pflag[j]){
                printf(" -");
            }
            else{
                printf(" %d",stu[i].p[j]);
            }
        }//輸出分數
        puts("");
    }
    return 0;
}