ZOJ Problem Set - 1007Numerical Summation of a Series

Numerical Summation of a Series

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Time Limit: 10 Seconds      Memory Limit: 32768 KB      Special Judge

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Produce a table of the values of the series

 

Equation 1

for the 2001 values of x, x= 0.000, 0.001, 0.002, ..., 2.000. All entries of the table must have an absolute error less than 0.5e-12 (12 digits of precision). This problem is based on a problem from Hamming (1962), when mainframes were very slow by today's microcomputer standards.

Input

This problem has no input.

Output

The output is to be formatted as two columns with the values of x andy(x) printed as in the C printf or the Pascal writeln.

printf("%5.3f %16.12f/n", x, psix )writeln(x:5:3, psix:16:12)

As an example, here are 4 acceptable lines out of 2001.

0.000   1.644934066848...0.500   1.227411277760...1.000   1.000000000000...2.000   0.750000000000

The values of x should start at 0.000 and increase by 0.001 until the line withx=2.000 is output.

Hint

The problem with summing the sequence in equation 1 is that too many terms may be required to complete the summation in the given time. Additionally, if enough terms were to be summed, roundoff would render any typical double precision computation useless for the desired precision.

To improve the convergence of the summation process note that

Equation 2

which implies y(1)=1.0. One can then produce a series fory(x) - y(1) which converges faster than the original series. This series not only converges much faster, it also reduces roundoff loss.

This process of finding a faster converging series may be repeated to produce sequences which converge more and more rapidly than the previous ones.

The following inequality is helpful in determining how may items are required in summing the series above.

Equation 3

 

 

这个题目之前AC过了，不过CW提起这个的时候忘了怎么写的了，现在重新看一次。写下来吧…

这个题目如果不是因为精确度被控制到小数点后第12位的话，纯粹是个水题。此公式如果强做的话，就算是要求10s也绝对会超时。因此必须考虑换种方法去计算。

ψ(1) = 1, ψ(x) = ψ(x) - ψ(1) + 1;因此，只要计算出ψ(x) - ψ(1)这部分值就OK了。

ψ(x) - ψ(1) = Σ(1/(k*(k+x))) - Σ(1/(k*(k + 1))) = Σ((1-x)/(k*(k+x)*(k+1))) = (1 - x)Σ(1/(k*(k+x)*(k+1)));

因此计算 Σ(1/(k*(k+x)*(k+1)))的值成为关键。我们可以先计算出K从1到某一个数的Σ(1/(k*(k+x)*(k+1)))的值，比如到k=20000；这样的话，我们必须估计出k > 20000的时候Σ(1/(k*(k+x)*(k+1)))的值，此时k>20000，（k+x）≈ k；则我们可以在程序开始之时先计算出k>20000,到k=100W甚至更高的时候Σ(1/(k*(k+x)*(k+1)))的估值，用temp保存起来，这个值会被利用

2001次…

#include <stdio.h>int main(void){ double p, i, j, temp = 0.0; for(j = 1000000; j > 20000; j --){ temp += 1.0 / (j * j * (j + 1)); } for(i = 0.000; i < 2.0005; i += 0.001){ p = temp; for(j = 20000; j >= 1; j--){ p += 1.0 / (j * (j + 1) * (j + i)); } p = (1 - i) * p; p += 1; printf("%5.3f %16.12f/n", i, p); } system("pause");}