上機考程式碼答案
每頁標題為超連結 可跳轉至zerojudge
-葉子齊-
#include <iostream>
using namespace std;
int main()
{
string m;
int n;
cin>>m>>n;
cout<<"I want "<<n<<" "<<m<<", thanks!";
return 0;
}出題者:林芊妘(教學長)
#include<iostream>
using namespace std;
int main()
{
string s;
while(getline(cin,s)) //用getline讓字串輸入不會被截斷
{
for(int i=0; i<s.length(); i++)
{ //無限判斷字串中的字源是否有符合上述的五個字
if(s[i]=='K' && s[i+1]=='e' && s[i+2]=='n' && s[i+3]=='s' && s[i+4]=='h' && s[i+5]=='i')
{
for(int j=i;j<=i+5;j++) s[j]='0';
}
if(s[i]=='Y' && s[i+1]=='o' && s[i+2]=='n' && s[i+3]=='e' && s[i+4]=='z' && s[i+5]=='u')
{
for(int j=i;j<=i+5;j++) s[j]='0';
}
if(s[i]=='c' && s[i+1]=='o' && s[i+2]=='n')
{
for(int j=i;j<=i+2;j++) s[j]='0';
if(s[i+3]=='c' && s[i+4]=='e' && s[i+5]=='r' && s[i+6]=='t')
{
for(int j=i+3;j<=i+6;j++) s[j]='0';
}
}
if(s[i]=='L' && s[i+1]=='e' && s[i+2]=='m' && s[i+3]=='o' && s[i+4]=='n')
{
for(int j=i;j<=i+4;j++) s[j]='0';
}
}
if(s=="0") break; //輸入0,則跳出迴圈結束
cout<<s<<endl;
}
}出題者:林芊妘(教學長)
#include <iostream>
using namespace std;
int main()
{
int Y=0,light=0,small=0,middle=0,large=0,sum=0,x=0;
while(cin>>Y>>light>>small>>middle>>large){
if(Y>=51 && Y<=69 && light<=7){
sum=small+(middle*2)+(large*3);
if(sum>=100)
cout<<"已經100顆辣!!"<<endl;
else{
x=100-sum;
cout<<"還有"<<x<<"顆,加油!!"<<endl;
}
}
else{cout<<"這裡根本沒有史萊姆!!"<<endl;}
x=0;
sum=0;
}
return 0;
}出題者:葉子齊(副社長)
#include <iostream>
using namespace std;
int main()
{
int D=0,X=0,EXP=0,Today_EXP=0,Total_EXP=0;
while(cin>>D>>X>>EXP){
for(int multiple=1;multiple<=D;multiple++){ //multiple倍數
Today_EXP=(X*EXP)*multiple;
Total_EXP+=Today_EXP;
}
cout<<Total_EXP<<endl;
Today_EXP=0;
Total_EXP=0;
}
return 0;
}
出題者:葉子齊(副社長)
解法一:
#include <iostream>
using namespace std;
int main()
{
int D=0,X=0,EXP=0;
while(cin>>D>>X>>EXP){
cout<<X*EXP*((D+1)*D/2)<<endl;
}
return 0;
}出題者:葉子齊(副社長)
解法二:
#include <iostream>
using namespace std;
int calculate(int m, int prices, int count){
if(count==m) return prices;
else{
count++;
return count*calculate(m, prices, count);
}
}
int main() {
int n, m;
cin >> n >> m;
int prices[101];
for (int i = 1; i <= n; i++) {
cin >> prices[i];
int count = 1;
int ans = calculate(m, prices[i], count);
if(ans > 10000) cout << 10000 << ' ';
else cout << ans << ' ';
}
cout << endl;
}出題者:蔡其霏(教學)
上機考程式碼答案
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