Topic`
No one hasn't grabbed red envelopes, right... Here is a record of N people giving red envelopes to each other and grabbing red envelopes. Please count their gains from grabbing red envelopes.
Input format:
Input the first line to give a positive integer N (≤104^4), that is, the total number of people participating in red envelope distribution and red envelope grab, then these people are numbered from 1 to N. In the next N lines, the i-th line gives the record of the red envelope issued by the person numbered i, in the following format: KN1 P1 …NK PK where K (0≤K≤20) It is the number of red envelopes sent. Ni is the number of the person who grabs the red envelope, and Pi (>0) is the amount of red envelopes (in cents). Note: For red envelopes issued by the same person, each person can only grab it once at most, and cannot be repeated.
Output format:
output each person’s serial number and income amount in descending order of income amount (in yuan, output 2 decimal places). Each person’s information is on one line with a space between the two numbers. If there is a tie in the amount of income, it will be output in descending order according to the number of red envelopes grabbed; if there is a tie, it will be output in increments according to the individual number.
Input example:
10
3 2 22 10 58 8 125
5 1 345 3 211 5 233 7 13 8 101
1 7 8800
2 1 1000 2 1000
2 4 250 10 320
6 5 11 9 22 8 33 7 44 10 55 4 2
1 3 8800
2 1 23 2 123
1 8 250
4 2 121 4 516 7 112 9 10
Output example:
1 11.63
2 3.63
8 3.63
3 2.11
7 1.69
6 -1.67
9 -2.18
10 -3.26
5 -3.26
4 -12.32
The qsort function is included in <stdlib.h>
Usage
Code:
#include <stdio.h>
#include <stdlib.h>
#include <malloc.h>
/* run this program using the console pauser or add your own getch, system("pause") or input loop */
struct peo{
int id;
int fmoney;
int smoney;
double jieguo;
int geshu;
};
int cmp(const void *a,const void *b){
struct peo s1 = *(struct peo*)a;
struct peo s2 = *(struct peo*)b;
if(s1.jieguo<s2.jieguo){
return 1;
}else if(s1.jieguo==s2.jieguo){
if(s1.geshu<s2.geshu){
return 1;
}else if(s1.geshu==s2.geshu){
if(s1.id>s2.id){
return 1;
}else if(s1.id==s2.id){
return 0;
}else{
return -1;
}
}else{
return -1;
}
}else{
return -1;
}
}
int main(int argc, char *argv[]) {
int i,j,n,m,a,b;
scanf("%d",&n);
struct peo p[10005] = {
0};
for(i=0;i<n;i++){
scanf("%d",&m); //发的红包个数
p[i].id = i+1;
for(j=0;j<m;j++){
scanf("%d%d",&a,&b); //a抢红包的人，b抢到的钱
p[a-1].smoney += b;
p[a-1].geshu++;
p[i].fmoney += b;
}
}
for(i=0;i<n;i++){
p[i].jieguo = p[i].smoney - p[i].fmoney;
}
qsort(p,n,sizeof(struct peo),cmp);
//输出
for(i=0;i<n;i++){
printf("%d %.2lf\n",p[i].id,p[i].jieguo/100);
}
return 0;
}