The 2014 ACM-ICPC Asia Mudanjiang Regional Contest - I.Information Entropy
2014-10-12 13:52
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题目大意:求解题目的第二个公式。
解题思路:直接求解第二个公式,只需注意下为0的时候,值为0即可,详见code。
题目来源:http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5359
code:
解题思路:直接求解第二个公式,只需注意下为0的时候,值为0即可,详见code。
题目来源:http://acm.zju.edu.cn/onlinejudge/showContestProblem.do?problemId=5359
code:
#include <iostream> #include <cstdio> #include <cstring> #include <cmath> using namespace std; const int eps = 1e-9; const int MAXN = 100+10; const int M = 5; int t,n,d; double p[MAXN]; char str[M]; int main(){ //freopen("input.txt","r",stdin); scanf("%d",&t); while(t--){ double ans=0; scanf("%d %s",&n,str); for(int i=0;i<n;++i){ scanf("%d",&d); p[i]=d*1.0/100; } if(str[0]=='b'){ for(int i=0;i<n;++i){ if(p[i]==0) continue; ans-=p[i]*log2l(p[i]+eps); } } if(str[0]=='n'){ for(int i=0;i<n;++i){ if(p[i]==0) continue; ans-=p[i]*logl(p[i]+eps); } } if(str[0]=='d'){ for(int i=0;i<n;++i){ if(p[i]==0) continue; ans-=p[i]*log10l(p[i]+eps); } } printf("%.11f\n",ans); } return 0; }
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