uva 10288 Coupons 数学期望

来源：互联网发布：sjcam sj5000x淘宝编辑：程序博客网时间：2024/06/13 23:42

计算期望可以从两个角度出发计算概率，第一个角度是直接计算，第二个角度是计算每增加一个的概率。
1+s+s^2+s^3+…+s^n n–>无穷，由等比数列前n项和公式为a1*(1-q^n)/(1-q);
q是分数，n是无穷，所以为1/(1-q)
1+2p+3p^2+….np^(n-1)，这个式子有两个特征，系数是递进的，抛开系数像是一个等比数列，那么我通过构造错位就可以构造出一个真正的等比数列。设这个式子为E，那么E-p*E就是一个等差数列了。

题目链接：http://ln1.vjudge.net/problem/481

#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;LL gcd(LL a, LL b){  return (b==0) ? a : gcd(b,a%b);}LL lcm(LL a, LL b){    return a / gcd(a,b) * b;}int main(){//    freopen("F:\\input.txt","r",stdin);//    freopen("F:\\output.txt","w",stdout);//    ios::sync_with_stdio(false);    LL lcmm[35];    ms(lcmm);    lcmm[1] = 1;    for(int i = 2; i <= 33; ++i){        lcmm[i] = lcm(lcmm[i-1],i);    }    int n;    int w1,w2;    double ans;    LL a,b;    LL gcdd;    LL ans1;    while(~scanf("%d",&n)){        if(n==1)            printf("1\n");        else if(n==2)            printf("3\n");        else{            a = 0;            for(int i = 1; i <= n; ++i){                a += lcmm[n]/i;            }            a = a*n;            b = lcmm[n];            ans1 = a/b;            a = a%b;            gcdd = gcd(a,b);            a /= gcdd;            b /= gcdd;            w1 = log10(ans1)+1;            w2 = log10(b)+1;            for(int i = 0; i < w1; ++i)                printf(" ");            printf(" %lld\n",a);            printf("%lld ",ans1);            for(int i = 0; i < w2; ++i)                printf("-");            printf("\n");            for(int i = 0; i < w1; ++i)                printf(" ");            printf(" %lld\n",b);        }    }    return 0;}

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