1108. Finding Average (20)
时间限制
400 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
The basic task is simple: given N real numbers, you are supposed to calculate their average. But what makes it complicated is that some of the input numbers might not be legal. A “legal” input is a real number in [-1000, 1000] and is accurate up to no more than 2 decimal places. When you calculate the average, those illegal numbers must not be counted in.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=100). Then N numbers are given in the next line, separated by one space.
Output Specification:
For each illegal input number, print in a line “ERROR: X is not a legal number” where X is the input. Then finally print in a line the result: “The average of K numbers is Y” where K is the number of legal inputs and Y is their average, accurate to 2 decimal places. In case the average cannot be calculated, output “Undefined” instead of Y. In case K is only 1, output “The average of 1 number is Y” instead.
Sample Input 1:
7
5 -3.2 aaa 9999 2.3.4 7.123 2.35
Sample Output 1:
ERROR: aaa is not a legal number
ERROR: 9999 is not a legal number
ERROR: 2.3.4 is not a legal number
ERROR: 7.123 is not a legal number
The average of 3 numbers is 1.38
Sample Input 2:
2
aaa -9999
Sample Output 2:
ERROR: aaa is not a legal number
ERROR: -9999 is not a legal number
The average of 0 numbers is Undefined
代码如下：
sscanf()可将字符串转为特定类型
sprintf()打印
#include<cstdio>
#include<cstring>
using namespace std;
int main(){
int n;
scanf("%d",&n);
int count=0;
char a[100],b[100];
double value,sum=0.00;
for(int i=0;i<n;++i){
scanf("%s",a);
sscanf(a,"%lf",&value);
sprintf(b,"%0.2lf",value);
int flag=0;
for(int j=0;j<strlen(a);++j){
if(a[j]!=b[j]){
flag=1;
}
}
if(flag||value<-1000||value>1000){
printf("ERROR: %s is not a legal number\n",a);
}else{
count++;
sum+=value;
}
}
if(count!=1&&count!=0){
printf("The average of %d numbers is %0.2lf",count,sum/count);
}else if(count==1){
printf("The average of 1 number is %0.2lf\n",sum);
}else{
printf("The average of 0 numbers is Undefined\n");
}
return 0;
}
参考了1108. Finding Average (20)-PAT甲级真题
1116. Come on! Let’s C (20)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
“Let’s C” is a popular and fun programming contest hosted by the College of Computer Science and Technology, Zhejiang University. Since the idea of the contest is for fun, the award rules are funny as the following:
\0. The Champion will receive a “Mystery Award” (such as a BIG collection of students’ research papers…). \1. Those who ranked as a prime number will receive the best award – the Minions (小黄人)! \2. Everyone else will receive chocolates.
Given the final ranklist and a sequence of contestant ID’s, you are supposed to tell the corresponding awards.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=10000), the total number of contestants. Then N lines of the ranklist follow, each in order gives a contestant’s ID (a 4-digit number). After the ranklist, there is a positive integer K followed by K query ID’s.
Output Specification:
For each query, print in a line “ID: award” where the award is “Mystery Award”, or “Minion”, or “Chocolate”. If the ID is not in the ranklist, print “Are you kidding?” instead. If the ID has been checked before, print “ID: Checked”.
Sample Input:
6
1111
6666
8888
1234
5555
0001
6
8888
0001
1111
2222
8888
2222
Sample Output:
8888: Minion
0001: Chocolate
1111: Mystery Award
2222: Are you kidding?
8888: Checked
2222: Are you kidding?
1116. Come on! Let’s C (20)
时间限制
200 ms
内存限制
65536 kB
代码长度限制
16000 B
判题程序
Standard
作者
CHEN, Yue
“Let’s C” is a popular and fun programming contest hosted by the College of Computer Science and Technology, Zhejiang University. Since the idea of the contest is for fun, the award rules are funny as the following:
\0. The Champion will receive a “Mystery Award” (such as a BIG collection of students’ research papers…). \1. Those who ranked as a prime number will receive the best award – the Minions (小黄人)! \2. Everyone else will receive chocolates.
Given the final ranklist and a sequence of contestant ID’s, you are supposed to tell the corresponding awards.
Input Specification:
Each input file contains one test case. For each case, the first line gives a positive integer N (<=10000), the total number of contestants. Then N lines of the ranklist follow, each in order gives a contestant’s ID (a 4-digit number). After the ranklist, there is a positive integer K followed by K query ID’s.
Output Specification:
For each query, print in a line “ID: award” where the award is “Mystery Award”, or “Minion”, or “Chocolate”. If the ID is not in the ranklist, print “Are you kidding?” instead. If the ID has been checked before, print “ID: Checked”.
Sample Input:
6
1111
6666
8888
1234
5555
0001
6
8888
0001
1111
2222
8888
2222
Sample Output:
8888: Minion
0001: Chocolate
1111: Mystery Award
2222: Are you kidding?
8888: Checked
2222: Are you kidding?
代码：
#include<iostream>
#include<vector>
#include<string>
using namespace std;
bool isprime(int value){
if(value==2) return true;
for(int i=2;i*i<=value;++i){
if(value%i==0){
return false;
}
}
return true;
}
int main(){
int n;
scanf("%d",&n);
int id;
vector<string> vec[10000];
for(int i=1;i<=n;++i){
scanf("%d",&id);
if(i==1){
vec[id].push_back("Mystery Award");
continue;
}
if(isprime(i)){
vec[id].push_back("Minion");
continue;
}
vec[id].push_back("Chocolate");
}
int query;
scanf("%d",&query);
vector<int> vec1;
for(int i=0;i<query;++i){
scanf("%d",&id);
vec1.push_back(id);
}
for(int i=0;i<vec1.size();++i){
printf("%04d",vec1[i]);
if(vec[vec1[i]].size()!=0){
if(vec[vec1[i]][0]!="1"){
cout<<": "<<vec[vec1[i]][0]<<endl;
vec[vec1[i]][0]="1";
}else{
cout<<": Checked"<<endl;
}
}else{
cout<<": Are you kidding?"<<endl;
}
}
return 0;
}
