求交错序列前N项和 #include <stdio.h> int main() { int numerator, denominator, flag, i, n; double item, sum; while (scanf("%d", &n) != EOF) { flag = 1; numerator = 1; denominator = 1; sum = 0; for (i = 1; i <= n; i++) { item = flag*1.0*numer
Sumdiv Time Limit: 1000MS Memory Limit: 30000K Total Submissions: 13959 Accepted: 3433 Description Consider two natural numbers A and B. Let S be the sum of all natural divisors of A^B. Determine S modulo 9901 (the rest of the division of S by 99
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