PAT Basic Level (1026~1030)

PAT 1026 程序运行时间

链接：:http://www.patest.cn/contests/pat-b-practise/1026

代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    int a,b,c,d,f,g;    scanf("%d%d",&a,&b);    c=b-a;    if(c/10%10>=5)        c=c/100+1;    else        c=c/100;    d=c/3600;    f=c%3600/60;    g=c%60;    if(d<10)        printf("0%d:",d);    else        printf("%d:",d);    if(f<10)        printf("0%d:",f);    else        printf("%d:",f);    if(g<10)        printf("0%d",g);    else        printf("%d",g);    return 0;}

PAT 1027 打印沙漏

链接：http://www.patest.cn/contests/pat-b-practise/1027

代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;int main(){    char c;    int n,m,high=0,width,r=0;    scanf("%d %c",&n,&c);    if(n<7)    {        printf("%c\n%d\n",c,n-1);        return 0;    }    m=n;    for(int i=1;;++high,i+=2)    {        if(i==1)        {            width=i;            m-=width;        }        else        {            width=i;            m-=(width*2);        }        if(m<(i+2)*2)        {            ++high;            r=m;            break;        }    }    //printf("%d %d\n",width,high);    for(int i=0;i<high;++i)    {        for(int j=0;j<i;++j)            printf(" ");        for(int j=0;j<width-i*2;++j)            printf("%c",c);        printf("\n");    }    int len=width/2;    for(int i=1;i<high;++i)    {        for(int j=0;j<len-i;++j)            printf(" ");        for(int j=0;j<i*2+1;++j)            printf("%c",c);        printf("\n");    }    printf("%d\n",r);    return 0;}

PAT 1028 人口普查

链接：http://www.patest.cn/contests/pat-b-practise/1028

代码：注意符合条件的人数可能为0

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<algorithm>using namespace std;struct person{    int y,m,d;    string name;} p,oldest,youngest;int st,en;int cal(struct person tmp){    return (tmp.y-1814)*365+tmp.m*30+tmp.d;}bool check(){    if(cal(p)<st||cal(p)>en) return false;    if(cal(p)>=cal(youngest))    {        youngest.name=p.name;        youngest.y=p.y;        youngest.m=p.m;        youngest.d=p.d;    }    if(cal(p)<=cal(oldest))    {        oldest.name=p.name;        oldest.y=p.y;        oldest.m=p.m;        oldest.d=p.d;    }    return true;}int main(){    int n,countx=0;    char c[10];    scanf("%d",&n);    youngest.y=1814;    youngest.m=9;    youngest.d=6;    oldest.y=2014;    oldest.m=9;    oldest.d=6;    st=9*30+6;    en=200*365+9*30+6;    for(int i=0; i<n; ++i)    {        scanf("%s%d/%d/%d",c,&p.y,&p.m,&p.d);        p.name=string(c);        if(check())        {            ++countx;        }    }    if(countx==0)        printf("0\n");    else        printf("%d %s %s\n",countx,oldest.name.c_str(),youngest.name.c_str());    return 0;}

PAT 1029 旧键盘

链接：http://www.patest.cn/contests/pat-b-practise/1029

代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<set>#include<algorithm>using namespace std;set<char> charSet;int main(){    bool flag;    char s1[100],s2[100];    scanf("%s%s",s1,s2);    //printf("%s\n%s\n",s1,s2);    int len1=strlen(s1);    int len2=strlen(s2);    for(int i=0,j=0; i<len1; ++i)    {        flag=false;        if(j==len2)        {            flag=true;        }        else        {            if(s1[i]==s2[j])                ++j;            else                flag=true;        }        if(flag)        {            if('a'<=s1[i]&&s1[i]<='z')                s1[i]=s1[i]+'A'-'a';            if((charSet.count(s1[i]))==0)            {                charSet.insert(s1[i]);                printf("%c",s1[i]);            }        }    }    printf("\n");    return 0;}

PAT 1030 完美数列

链接：http://www.patest.cn/contests/pat-b-practise/1030

代码：

#include<iostream>#include<string>#include<cstdio>#include<cstring>#include<set>#include<algorithm>using namespace std;double v[100005];int main(){    int n,maxx=0;    double p;    scanf("%d%lf",&n,&p);    for(int i=0;i<n;++i)        scanf("%lf",&v[i]);    sort(v,v+n);    for(int i=0;i<n; i++)        for(int j=i+maxx-1; j<n; j++)        {            if(v[i]*p<v[j])                break;            if(j-i+1>maxx)                maxx =j-i+1;        }    printf("%d\n",maxx);    return 0;}