PAT(basic level) 1012 数字分类(20)
2016-11-16 20:19
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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:
输出样例1:
输入样例2:
输出样例2:
#include <iostream>
using namespace std;
int main()
{
int n;
scanf("%d", &n);
int a
; //存输入的N个数
for(int i = 0; i < n; i ++)
scanf("%d", &a[i]);
int k = 0, sum0 = 0, sum1 = 0, count2 = 0, sum3 = 0, count3 = 0, max = -1, count0 = 0, count1 = 0;
for(int i = 0; i < n; i ++){
if(a[i]%5 == 0 && a[i]%2 == 0) {sum0 += a[i]; count0 ++; continue;}
if(a[i]%5 == 1) {
if(k == 0) {sum1 += a[i];count1 ++; k = 1;}
else {sum1 -= a[i]; count1 ++; k = 0;}
continue;
}
if(a[i]%5 == 2) {count2 ++; continue;}
if(a[i]%5 == 3) {sum3 += a[i]; count3 ++; continue;}
if(a[i]%5 == 4) {if(max < a[i]) max = a[i]; continue;}
}
double avg = (double)sum3/(double)count3;
if(count0 != 0) printf("%d ",sum0); else printf("N ");
if(count1 != 0) printf("%d ",sum1); else printf("N ");
if(count2 != 0) printf("%d ",count2); else printf("N ");
if(count3 != 0) printf("%.1f ",avg); else printf("N ");
if(max != -1) printf("%d",max); else printf("N");
return 0;
}
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>
using namespace std;
int main()
{
int n;
scanf("%d", &n);
int a
; //存输入的N个数
for(int i = 0; i < n; i ++)
scanf("%d", &a[i]);
int k = 0, sum0 = 0, sum1 = 0, count2 = 0, sum3 = 0, count3 = 0, max = -1, count0 = 0, count1 = 0;
for(int i = 0; i < n; i ++){
if(a[i]%5 == 0 && a[i]%2 == 0) {sum0 += a[i]; count0 ++; continue;}
if(a[i]%5 == 1) {
if(k == 0) {sum1 += a[i];count1 ++; k = 1;}
else {sum1 -= a[i]; count1 ++; k = 0;}
continue;
}
if(a[i]%5 == 2) {count2 ++; continue;}
if(a[i]%5 == 3) {sum3 += a[i]; count3 ++; continue;}
if(a[i]%5 == 4) {if(max < a[i]) max = a[i]; continue;}
}
double avg = (double)sum3/(double)count3;
if(count0 != 0) printf("%d ",sum0); else printf("N ");
if(count1 != 0) printf("%d ",sum1); else printf("N ");
if(count2 != 0) printf("%d ",count2); else printf("N ");
if(count3 != 0) printf("%.1f ",avg); else printf("N ");
if(max != -1) printf("%d",max); else printf("N");
return 0;
}
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