1、問題

A(n) = n / (2 * n + 1)
B1 = 2 + A1;
B2 = 2 + A1 * (2 + A2);
B3 = 2 + A1 * (2 + A2 * (2 + A3));
....以此類推，求B(n)

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2、代碼實(shí)現(xiàn)

#include <stdio.h>/** A(n) = n / (2 * n + 1) B1 = 2 + A1; B2 = 2 + A1 * (2 + A2); B3 = 2 + A1 * (2 + A2 * (2 + A3)); ....以此類推，求B() **/ float A(float n) {if (n < 0)return 0;float result = n / (2 * n + 1);return result; }//非遞歸實(shí)現(xiàn) float B(float n) {if (n < 0)return 0;float sum = 1;for (int i = n; i >= 1; --i) {sum = sum * A(i) + 2;}return sum; }//遞歸實(shí)現(xiàn) float recursion_B(float n) {if (n < 0)return 0;static float sum = 1;if (n == 0)return sum;else {sum = sum * A(n) + 2;recursion_B(n - 1); } }int main() {for (int i = 0; i < 20; i++)printf("B(%d) is %f\n", i, B(i));printf("recursion_B(10) is %f\n", recursion_B(10)); }
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3、運(yùn)行結(jié)果

B(0) is 1.000000 B(1) is 2.333333 B(2) is 2.800000 B(3) is 2.990476 B(4) is 3.073016 B(5) is 3.109957 B(6) is 3.126829 B(7) is 3.134643 B(8) is 3.138300 B(9) is 3.140024 B(10) is 3.140842 B(11) is 3.141232 B(12) is 3.141419 B(13) is 3.141509 B(14) is 3.141552 B(15) is 3.141573 B(16) is 3.141583 B(17) is 3.141588 B(18) is 3.141591 B(19) is 3.141592 recursion_B(10) is 3.140842

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