PATB1012

最新推荐文章于 2025-02-14 14:07:05 发布

MurrayCod

最新推荐文章于 2025-02-14 14:07:05 发布

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文章标签： PAT

本文链接：https://blog.youkuaiyun.com/m0_37483549/article/details/70161603

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A1 n%10==0 求和

A2 n%5==1 && n%10!=1求和减去 n%10==1求和

A3 count（n%5==2）

A4 avg(n%5==3) double 输出格式为%0.1d

A4 max(n%5==4)

#include <cstdio>
#define MAXN 1000
int main()
{    int a[MAXN],n,i,a1=0,a3=0,a2=0,count2=0,count3=0,max=0,flag=1;
    double avg3=0;
    scanf("%d",&n);
    for(i=1;i<=n;i++){
        scanf("%d",&a[i]);}
    for(i=1;i<=n;i++){

        if(a[i]%10==0) a1=a1+a[i];

        else if(a[i]%5==1){
            if(flag%2==1){a2=a2+a[i];flag++;}
            else{a2=a2-a[i];flag++;}}

        else if(a[i]%5==2) count2++;

        else if(a[i]%5==3) {a3=a3+a[i];count3++;}

        else if(a[i]%5==4) if(a[i]>max) max=a[i];} 
        avg3= a3/double(count3);//判断情况

    if(a1) printf("%d ",a1);
    else printf("N ");

    if(a2!=0) printf("%d ",a2);
    else printf("N ");

    if(count2) printf("%d ",count2);
    else printf("N ");

    if(avg3) printf("%.1f ",avg3);
    else printf("N ");

    if(max) printf("%d",max);
    else printf("N");
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