Created
October 28, 2009 15:56
-
-
Save anonymous/220560 to your computer and use it in GitHub Desktop.
This file contains hidden or bidirectional Unicode text that may be interpreted or compiled differently than what appears below. To review, open the file in an editor that reveals hidden Unicode characters.
Learn more about bidirectional Unicode characters
#include <stdio.h> | |
#include <string.h> | |
#include <stdlib.h> | |
int main() { | |
int i,j,N,K,cas=1; | |
double ans,tmp; | |
// freopen("test.in", "r", stdin); | |
// freopen("test.out", "w", stdout); | |
while(scanf("%d%d", &N, &K) != EOF) { | |
if(N == 0 && K == 0) | |
break; | |
if(N == 1) | |
printf("Case %d: %.4f\n", cas++, 0.0); | |
else if(K * 2 + 1 >= N) | |
printf("Case %d: %.4f\n", cas++, (double)N); | |
else { | |
ans = 0.0; | |
for(i=N-1,j=N-(K*2+1),tmp=1.0; i>=N-K*2; i--,j--) { | |
tmp *= (double)j / i; | |
} | |
ans += (1.0 - tmp) * (N - K*2); | |
// printf("%f %f\n", tmp, ans); | |
for(i=N-K*2; i<=N-(K+1); i++) { | |
if(i - K*2 == 0) { | |
for(i=1,tmp=1.0; i<=K*2; i++) { | |
tmp *= (double)i / (N - K*2 + i - 1); | |
} | |
i--; | |
if(i >= N - K*2 && i <= N - (K + 1)) | |
ans += (1.0 - tmp) * 2.0; | |
} | |
else { | |
tmp *= (double)i / (i - K*2); | |
ans += (1.0 - tmp) * 2.0; | |
} | |
// printf("%f %f\n", tmp, ans); | |
} | |
printf("Case %d: %.4f\n", cas++, ans); | |
} | |
} | |
return 0; | |
} | |
If you use the long double instead double, your code will get Accepted!!
Sign up for free
to join this conversation on GitHub.
Already have an account?
Sign in to comment
Judge verdict: WA