[UVA] 11348 - Exhibition
I - Exhibition
Time Limit: 1 sec
Memory Limit: 16MB
Some friends have the same hobby, they are collecting stamps. Once upon a time they decided to make an exhibition. Exhibition brought them some money and now they do not know how to divide their income. They decided to divide their money according this rule: "The percent of whole income that i-th friend will get is equal to the part of his unique stamp's type."
The stamp type is called unique if and only if this type of stamps of owned only by one person.
INPUT:
The first line contains integer K (0 < K <= 100), it is number of tests. Each test case is described by positive integer N (0 < N <= 50), it`s the number of friends. Next goes N lines with integers. Each line corresponds one friend stamp collection. The first integer on the line is M - the number of stamps owned by a person (0 < M <= 50). Next goes M integers A (0 <= A <= 10000) - types of stams.OUTPUT:
For each test case out line formatter like this: "Case i: a1%
a2% a3% ... an%". Where "i"
is a test number, and "ai" percent of income that goes
to i-th friend.
SAMPLE INPUT:
1
3
3 1 2 3
2 4 5
3 4 2 6
SAMPLE OUTPUT:
Case 1: 50.000000% 25.000000% 25.000000%
Problem setters: Aleksej Viktorchik, Leonid Shishlo.
Huge Easy Contest #1
題目不難，但有陷阱，一個人可能擁有多個同一種郵票。
因此在獨特郵票的判定上，要特別小心，只要只有一個人有，他仍是一張獨特的郵票。
#include <stdio.h>
#include <algorithm>
using namespace std;
int main() {
int testcase, cases = 0;
int A[50][50], Am[50];
int n, i, j;
scanf("%d", &testcase);
while(testcase--) {
scanf("%d", &n);
int mark[10005] = {};
for(i = 0; i < n; i++) {
scanf("%d", &Am[i]);
for(j = 0; j < Am[i]; j++)
scanf("%d", &A[i][j]);
sort(A[i], A[i]+Am[i]);
mark[A[i][0]]++;
for(j = 1; j < Am[i]; j++)
if(A[i][j] != A[i][j-1])
mark[A[i][j]]++;
}
int unique = 0;
for(i = 0; i <= 10000; i++)
unique += (mark[i] == 1);
printf("Case %d:", ++cases);
for(i = 0; i < n; i++) {
int cnt = 0;
cnt += (mark[A[i][0]] == 1);
for(j = 1; j < Am[i]; j++) {
if(A[i][j] != A[i][j-1])
cnt += (mark[A[i][j]] == 1);
}
printf(" %.6lf%%", (double)cnt*100.0/unique);
}
puts("");
}
return 0;
}