UVa 10883 - Supermean （杨辉三角 通过取对数解决大数除大数）

4
1
10.4
2
1.0 2.2
3
1 2 3
5
1 2 3 4 5

Case #1: 10.400
Case #2: 1.600
Case #3: 2.000
Case #4: 3.000

题意：

给出n个数，每相邻两个数求平均数，得到n-1个数，然后再继续这有求得到n-2个，一直到最后1个数。求这个数

手算一下就知道是二项式系数

ans = ∑C(n-1,i) / (2^(n-1))

但是这里范围特别大，所以需要取对数来算

原理：e^log(e,a) = a?

#include <cstdio>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
#include <string>
#include <map>
#include <cmath>
#include <queue>
#include <set>

using namespace std;

//#define WIN
#ifdef WIN
typedef __int64 LL;
#define iform "%I64d"
#define oform "%I64dn"
#define oform1 "%I64d"
#else
typedef long long LL;
#define iform "%lld"
#define oform "%lldn"
#define oform1 "%lld"
#endif

#define S64I(a) scanf(iform,&(a))
#define P64I(a) printf(oform,(a))
#define P64I1(a) printf(oform1,(a))
#define REP(i,n) for(int (i)=0; (i)<n; (i)++)
#define REP1(i,n) for(int (i)=1; (i)<=(n); (i)++)
#define FOR(i,s,t) for(int (i)=(s); (i)<=(t); (i)++)

const int INF = 0x3f3f3f3f;
const double eps = 1e-9;
const double PI = (4.0*atan(1.0));

int main() {
    int T;

    scanf("%d",&T);
    for(int kase=1; kase<=T; kase++) {
        int n;
        double lnc = 0;
        double ans = 0;
        scanf("%d",&n);
        double ln2 = (n-1) * log(2.0);
        for(int i=0; i<n; i++) {
            double val;
            scanf("%lf",&val);
            if(val > 0) ans += exp(log(val) + lnc - ln2);
            else ans -= exp(log(-val) + lnc - ln2);
            lnc = lnc + log(n-i-1) - log(i+1);
        }
        printf("Case #%d: %.3lfn",kase,ans);
    }

    return 0;
}