题目描述
解题思路
对于A1到A5，我们用5个函数分别求。
在下面的代码中，用
find表示是否查找到出现过该种类型的数字，出现过赋值为1，没有出现为0.具体实现看下面的代码
代码
#include <stdio.h> #include <stdlib.h> int find = 0; int f1(int* a, int n) { find = 0; int i = 0,sum=0; for (; i < n; i++) { if (a[i] % 5 == 0&&a[i]%2==0) { find = 1; sum += a[i]; } } return sum; } int f2(int* a, int n) { find = 0; int i = 0, sum = 0, x = 1; for (; i < n; i++) { if (a[i] % 5 == 1) { find = 1; sum += x * a[i]; x = -1*x; } } return sum; } int f3(int* a, int n) { find = 0; int i = 0, sum = 0; for (; i < n; i++) { if (a[i] % 5 == 2 ) { find = 1; sum++; } } return sum; } double f4(int* a, int n) { find = 0; int i = 0, sum = 0,count=0; for (; i < n; i++) { if (a[i] % 5 == 3 ) { find = 1; sum += a[i]; count++; } } return (double)sum/count; } int f5(int* a, int n) { find = 0; int i = 0, max = 0; for (; i < n; i++) { if (a[i] % 5 == 4) { find = 1; if(max<a[i]) max = a[i]; } } return max; } int main() { int n,i; scanf("%d", &n); int* a = (int*)malloc(n * sizeof(int)); if (a == NULL) { return 0; } for (i = 0; i < n; i++) { scanf("%d", a + i); } int ret=f1(a, n); if (find == 0) { printf("N "); } else { printf("%d ",ret); } ret=f2(a, n); if (find == 0) { printf("N "); } else { printf("%d ", ret); } ret = f3(a, n); if (find == 0) { printf("N "); } else { printf("%d ", ret); } double b=f4(a, n); if (find == 0) { printf("N "); } else { printf("%.1lf ", b); } ret = f5(a, n); if (find == 0) { printf("N "); } else { printf("%d ", ret); } free(a); a=NULL: return 0; }
