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/*给定一系列正整数，请按要求对数字进行分类，并输出以下5个数字：A1 = 能被5整除的数字中所有偶数的和；A2 = 将被5除后余1的数字按给出顺序进行交错求和，即计算n1-n2+n3-n4...；A3 = 被5除后余2的数字的个数；A4 = 被5除后余3的数字的平均数，精确到小数点后1位；A5 = 被5除后余4的数字中最大数字。输入格式：每个输入包含1个测试用例。每个测试用例先给出一个不超过1000的正整数N，随后给出N个不超过1000的待分类的正整数。数字间以空格分隔。输出格式：对给定的N个正整数，按题目要求计算A1~A5并在一行中顺序输出。数字间以空格分隔，但行末不得有多余空格。若其中某一类数字不存在，则在相应位置输出“N”。输入样例1：13 1 2 3 4 5 6 7 8 9 10 20 16 18输出样例1：30 11 2 9.7 9输入样例2：8 1 2 4 5 6 7 9 16输出样例2：N 11 2 N 9*/#include <iostream>#include <algorithm>#include <cstdio>#include <cstring>using namespace std;int main() {int n, num, sum1, sum2, sum3, sum4, sum5, cnt = 0;scanf("%d", &n);sum1 = 0;sum2 = 0;sum3 = 0;sum4 = 0;sum5 = -1;//加了一个mark标记就过了，可能其中一个测试数据中间值为0了吧int flag = 1, mark1 = 0, mark2 = 0, mark3 = 0, mark4 = 0, mark5 = 0;while(n--) {scanf("%d", &num);if(num % 5 == 0 && num % 2 == 0) {mark1 = 1;sum1 += num;}if(num % 5 == 1) {mark2 = 1;sum2 += flag * num;flag = -flag;}if(num % 5 == 2) {mark3 = 1;sum3++;}if(num % 5 == 3) {mark4 = 1;sum4 += num;cnt++;}if(num % 5 == 4) {mark5 = 1;if(num > sum5) {sum5 = num;}}}if(mark1 == 0) {printf("N ");}else {printf("%d ", sum1);}if(mark2 == 0) {printf("N ");}else {printf("%d ", sum2);}if(mark3 == 0) {printf("N ");}else {printf("%d ", sum3);}if(mark4 == 0) {printf("N ");}else {printf("%.1f ", sum4 * 1.0/ cnt);}if(mark5 == 0) {printf("N");}else {printf("%d", sum5);}return 0;}