uva 1639 Candy 大数的对数处理 数学期望

1. 当排列组合数或者幂很大时可以利用对数计算，之后再用exp还原，保证一定的精度。
2. 数学期望是每一个可能的值和相应的概率的乘积和，没有可能值可以设。
3. 仔细读题，吃完最后一个糖果后不知道是否已经吃完，所以需要再选一次。
4. %f用来输入float,输出float double.
5. %lf 用来输入double,输出long double.

题目链接：https://uva.onlinejudge.org/index.php?option=com_onlinejudge&Itemid=8&page=show_problem&category=825&problem=4514&mosmsg=Submission+received+with+ID+17971092

#include<cstdio>#include<iostream>#include<sstream>#include<cstdlib>#include<cmath>#include<cctype>#include<string>#include<cstring>#include<algorithm>#include<stack>#include<queue>#include<set>#include<map>#include<ctime>#include<vector>#include<fstream>#include<list>using namespace std;#define ms(s) memset(s,0,sizeof(s))typedef unsigned long long ULL;typedef long long LL;const int INF = 0x3fffffff;const int maxn = 400010;long double logF[maxn];void make_logF(){    logF[0] = 0;    for(int i = 1; i <= maxn; ++i)        logF[i] = logF[i-1]+log(i);}long double getC(int n, int r){    return logF[n] - logF[n-r] - logF[r];}int main(){    freopen("F:\\input.txt","r",stdin);//    freopen("F:\\output.txt","w",stdout);//    ios::sync_with_stdio(false);    int n;    double p;    double ans;    long double t1,t2,c;    int cas = 1;    make_logF();    while(~scanf("%d%lf",&n,&p)){        ans = 0;        for(int i = 0; i <= n; ++i){            c = getC(2*n-i,n);            t1 = c + (n+1)*log(p)+(n-i)*log(1-p);            t2 = c + (n+1)*log(1-p)+(n-i)*log(p);            ans += i*(exp(t1)+exp(t2));        }        printf("Case %d: %.6f\n",cas,ans);        cas++;    }    return 0;}