UVA - 10886 Standard Deviation （标准差）（数论）

题意：下面是一个随机数发生器。输入seed的初始值，你的任务是求出它得到的前n个随机数标准差，保留小数点后5位(1<=n<=10000000,0<=seed<2⁶⁴)。

分析：方差的式子最好化简一下再计算，否则按部就班，提前算出平均数，代入计算会产生精度误差导致WA。

化简方法：

将上式全部展开，并合并同类项，可得∑(xi²) / n - m²，m是平均数。求标准差，开根号即可。

#pragma comment(linker, "/STACK:102400000, 102400000")

#include<cstdio>

#include<cstring>

#include<cstdlib>

#include<cctype>

#include<cmath>

#include<iostream>

#include<sstream>

#include<iterator>

#include<algorithm>

#include<string>

#include<vector>

#include<set>

#include<map>

#include<stack>

#include<deque>

#include<queue>

#include<list>

#define Min(a, b) ((a < b) ? a : b)

#define Max(a, b) ((a < b) ? b : a)

const double eps = 1e-8;

inline int dcmp(double a, double b) {

    if(fabs(a - b) < eps)  return 0;

    return a < b ? -1 : 1;

}

typedef long long LL;

typedef unsigned long long ULL;

const int INT_INF = 0x3f3f3f3f;

const int INT_M_INF = 0x7f7f7f7f;

const LL LL_INF = 0x3f3f3f3f3f3f3f3f;

const LL LL_M_INF = 0x7f7f7f7f7f7f7f7f;

const int dr[] = {0, 0, -1, 1, -1, -1, 1, 1};

const int dc[] = {-1, 1, 0, 0, -1, 1, -1, 1};

const int MOD = 1e9 + 7;

const double pi = acos(-1.0);

const int MAXN = 1e7 + 10;

const int MAXT = 10000 + 10;

using namespace std;

ULL seed;

long double gen(){

    static const long double Z = (long double)1.0 / (1LL << 32);

    seed >>= 16;

    seed &= (1ULL << 32) - 1;

    seed *= seed;

    return seed * Z;

}

int main(){

    int T;

    scanf("%d", &T);

    int kase = 0;

    while(T--){

        int n;

        scanf("%d%llu", &n, &seed);

        double sum1 = 0;

        double sum2 = 0;

        for(int i = 0; i < n; ++i){

            double tmp = gen();

            sum1 += tmp * tmp;

            sum2 += tmp;

        }

        sum1 /= n;

        sum2 /= n;

        printf("Case #%d: %.5lf\n", ++kase, sqrt(sum1 - sum2 * sum2));

    }

    return 0;

}

巴特西

UVA - 10886 Standard Deviation （标准差）（数论）

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