imported>ardiefox at 11:22, 22 September 2014

2014-09-22T11:22:23Z

New page

https://icpcarchive.ecs.baylor.edu/external/65/6542.pdf

솔직히 이 문제 너무 쉬웠다고밖에는.......

== 황현 ==
#include <stdio.h>
#include <math.h>

int main() {
int T, c, x, y, i;
double p;

scanf("%d", &T);
for (i = 0; i < T; ++i) {
x = y = 0;
p = .0;
scanf("%d", &c);
getc(stdin);
switch (getc(stdin)) {
case 'Y':
x += 3;
y += 10;
case 'Z':
x += 3;
y += 10;
case 'E':
x += 3;
y += 10;
case 'P':
x += 3;
y += 10;
case 'T':
x += 3;
y += 10;
case 'G':
x += 3;
y += 10;
case 'M':
x += 3;
y += 10;
case 'K':
x += 3;
y += 10;
getc(stdin);
case 'B':
break;
}
getc(stdin);
y -= x;
printf("Case #%d: %.2lf%%\n", i + 1, round((1.0 - (pow(5, x) / pow(2, y))) * 10000.0) / 100.0);
}

return 0;
}

ACM ICPC/Problems/6542 - Revision history

imported>ardiefox at 11:22, 22 September 2014