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The group agrees in advance to share expenses equally, but it is not practical to have them share every expense as it occurs. So individuals in the group pay for particular things, like meals, hotels, taxi rides, plane tickets, etc. After the trip, each student‘s expenses are tallied and money is exchanged so that the net cost to each is the same, to within one cent. In the past, this money exchange has been tedious and time consuming. Your job is to compute, from a list of expenses, the minimum amount of money that must change hands in order to equalize (within a cent) all the students‘ costs.
Input
Standard input will contain the information for several trips. The information for each trip consists of a line containing a positive integer, n, the number of students on the trip, followed by n lines of input, each containing the amount, in dollars and cents, spent by a student. There are no more than 1000 students and no student spent more than $10,000.00. A single line containing 0 follows the information for the last trip.Output
For each trip, output a line stating the total amount of money, in dollars and cents, that must be exchanged to equalize the students‘ costs.Sample Input
3 10.00 20.00 30.00 4 15.00 15.01 3.00 3.01 0
Sample Output
$10.00 $11.99
Source: Waterloo
Local Contest Jan. 31, 1999
几点注意：
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1 /* 2 #include <stdio.h> 3 #include <string.h> 4 5 float cost[100]; 6 int cost100[100]; 7 8 void sort(int num); 9 10 int main(void) 11 { 12 int num,i,sum=0; 13 int avg,last; 14 int change_sum=0; 15 double change; 16 17 scanf("%d",&num); 18 while(num != 0){ 19 for(i=0;i<num;i++){ 20 scanf("%f",&cost[i]); 21 cost100[i] = 100*cost[i] + 0.5; 22 sum += cost100[i]; 23 } 24 avg = sum/num; 25 last = sum%num; 26 sort(num); 27 for(i=0;i<num;i++){ 28 if(i<last) 29 cost100[i] = cost100[i] - avg - 1; 30 else 31 cost100[i] = cost100[i] - avg; 32 if(cost100[i] > 0) 33 change_sum += cost100[i]; 34 } 35 36 change = ((double)change_sum)/100; 37 printf("$%.2f\n",change); 38 39 memset(cost,0,num*sizeof(float)); 40 memset(cost100,0,num*sizeof(int)); 41 42 sum = 0; 43 change_sum = 0; 44 45 scanf("%d",&num); 46 } 47 48 //system("PAUSE"); 49 return 0; 50 } 51 void sort(int num) 52 { 53 int i,j; 54 for(i=0;i<num;i++){ 55 for(j=0;j<num;j++){ 56 if(cost100[j]<cost100[j+1]){ 57 int temp = cost100[j]; 58 cost100[j] = cost100[j+1]; 59 cost100[j+1] = temp; 60 } 61 } 62 } 63 64 } 65 */ 66 67 68 #include <stdio.h> 69 #define num 1005 70 double s[num]; 71 int main() 72 { 73 int n, t, i; 74 while (scanf("%d", &n) == 1 && n) 75 { 76 double sum = 0, resultH = 0, resultL = 0; 77 for (i = 0; i < n; i++) 78 { 79 scanf("%lf", &s[i]); 80 sum += s[i]; 81 } 82 sum /= n; 83 84 for (i = 0; i < n; i++) 85 if (s[i] < sum) 86 { 87 resultL += (int)((sum - s[i])*100) / 100.0; 88 } 89 else 90 resultH += (int)((s[i] - sum)*100) / 100.0; 91 printf("$%0.2lf\n", resultL < resultH ? resultH:resultL); 92 } 93 return 0; 94 }
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原文地址：http://www.cnblogs.com/yushuo1990/p/4314170.html
