求奇数分之一序列前N项和 #include <stdio.h> int main() { int denominator, i, n; double item, sum; while (scanf("%d", &n) != EOF) { denominator = 1; sum = 0; for (i = 1; i <= n; i++) { item = 1.0/denominator; sum = sum+item; denominator = denomi
求N分之一序列前N项和 #include <stdio.h> int main() { int i, n; double item, sum; while (scanf("%d", &n) != EOF) { sum = 0; for (i = 1; i <= n; i++) { item = 1.0/i; sum = sum+item; } printf("sum = %f\n", sum); } return 0; }
练习2-13 求N分之一序列前N项和 (15 分) 输入在一行中给出一个正整数N. 输出格式: 在一行中按照“sum = S”的格式输出部分和的值S,精确到小数点后6位.题目保证计算结果不超过双精度范围. 输入样例: 6 输出样例: sum = 2.450000 #include <stdio.h> #include <stdlib.h> /* run this program using the console pauser or add your own getch, syst
求阶乘序列前N项和 #include <stdio.h> double fact(int n); int main() { int i, n; double item, sum; while (scanf("%d", &n) != EOF) { sum = 0; if (n <= 12) { for (i = 1; i <= n; i++) { item = fact(i); sum = sum + item; } } printf("%.0f
求平方根序列前N项和 #include <stdio.h> #include <math.h> int main() { int i, n; double item, sum; while (scanf("%d", &n) != EOF) { sum = 0; for (i = 1; i <= n; i++) { item = sqrt(i); sum = sum+item; } printf("sum = %.2f\n", s
后一个分数的分子=前一个分数的分子+分母,后一个分数的分母=前一个分数的分子,循环个20次就有结果.注意,假设分子为a,分母为b,虽然 a = a + b, 但此时a已经变成 a+b 了,所以再给b重新赋值的时候,得是 (a+b)-b 才能等于原分母b,所以重新赋值时就得写成 a-b 方法一 from fractions import Fraction def fibonacci(n): a, b = 1, 2 res = [1] i = 1 while i < n: a, b = b, a+b
A - Farey Sequence Time Limit:1000MS Memory Limit:65536KB 64bit IO Format:%I64d & %I64u Submit Status Practice POJ 2478 Description The Farey Sequence Fn for any integer n with n >= 2 is the set of irreducible rational numbers a/b with 0 &l
