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/*
一瓶可乐2元,2个空瓶换一瓶可乐,有X元钱,可喝到几瓶可乐..用编程算出.
*/
#include<iostream>
using namespace std;
int main()
{
int x = 0;
while(cout<<"Input Money: ",cin>>x)
{
int buyNum = x/2;
int redeemNum = buyNum/2;
cout<<x<<" mony can get "<<buyNum+redeemNum<<" bottle Coca."<<endl<<endl;
}
return 0;
}
#include<stdio.h>
int main()
{
int money;
printf("Money:");
scanf("%d",&money);
printf("bottles:%d\n",(money%2)?(money-1):money);
return 0;
}
#include<stdio.h>
int main()
{
int money;
printf("Money:");
scanf("%d",&money);
printf("bottles:%d\n",(money%2)?money:(money-1));
return 0;
}
#include <iostream>
using namespace std;
int BottleNum(int i)
{
int num=0;
for (int n=1;;n++)
{
if (i%2==0)
{
i=i/2;
num+=i;
}
else
{
i=(i-1)/2;
num+=i;
i++;
if (i==1)
{
break;
}
}
}
return num+1;//要是老板愿意再借你一瓶,你就可以拿着一个空瓶再喝一杯了
}
void main()
{
int x=19,j;
j=BottleNum(x);
cout<<j<<endl;
}
#include <stdio.h>
int main( void )
{
int money = 0;
int bottle = 0;
scanf("%d", &money);
bottle = money / 2;
printf("bottleTotal = %d\n", bottle * 2);
return 0;
}
#include <stdio.h>
/*
** 一瓶可乐2元,2个空瓶换一瓶可乐,有X元钱,可喝到几瓶可乐..用编程算出.
*/
int GetColeNum(int n){
if(n%2==1)return GetColeNum(n-1);
int num = n;
int temp=n;
while (n > 1) {
num += n / 2;
n = n/2 + n%2;
}
return num-temp;
}
int main(){
int x=0;
while (scanf("%d", &x) != EOF) {
printf("F(%d) = %d\n", x, GetColeNum(x));
}
return 0;
}