#12390: C++ CPP


swsamuel (世外唐悦)


#include <iostream>
using namespace std;

int main() {
    int AMOUNT;  

    while (cin >> AMOUNT) {
    if (AMOUNT >= 0 && AMOUNT <= 10)    
        cout << AMOUNT * 6 << endl;   
    else if (AMOUNT >= 11 && AMOUNT <= 20)
        cout << 60 + (AMOUNT - 10) * 2 << endl;    
    else if (AMOUNT >= 21 && AMOUNT <= 40)
        cout << 80 + (AMOUNT - 20) << endl;    
    else if (AMOUNT > 40)
        cout << 100 << endl;    
    }
}
代码如上;
不难,只是用多重if语句;
AC的,没错。
#13080: Re:C++ CPP


fdhs107_KonChin_Shih (Konchin)


#include 
using namespace std;

int main() {
    int AMOUNT;  

    while (cin >> AMOUNT) {
    if (AMOUNT >= 0 && AMOUNT <= 10)    
        cout << AMOUNT * 6 << endl;   
    else if (AMOUNT >= 11 && AMOUNT <= 20)
        cout << 60 + (AMOUNT - 10) * 2 << endl;    
    else if (AMOUNT >= 21 && AMOUNT <= 40)
        cout << 80 + (AMOUNT - 20) << endl;    
    else if (AMOUNT > 40)
        cout << 100 << endl;    
    }
}
代码如上;
不难,只是用多重if语句;
AC的,没错。

可是你的變數好難判讀喔(苦笑

#13613: Re:C++ CPP


snakeneedy (蛇~Snake)


分享不用 && 的寫法

#include <cstdio>

int main() {
  unsigned n;
  while (scanf("%u", &n) != EOF) {
    if (n >= 40) puts("100");
    else if (n >= 21) printf("%u\n", 80 + (n - 20));
    else if (n >= 11) printf("%u\n", 60 + (n - 10) * 2);
    else printf("%u\n", n * 6);
  }
}

 

#17258: Re:C++ CPP


anniechang2005 (啵的四葉草)


這樣很好,但最後一次可以直接 else cout<<100<<endl;
(小分享霸了,還請多多指教~

#include using namespace std; int main() { int AMOUNT; while (cin >> AMOUNT) { if (AMOUNT >= 0 && AMOUNT <= 10) cout << AMOUNT * 6 << endl; else if (AMOUNT >= 11 && AMOUNT <= 20) cout << 60 + (AMOUNT - 10) * 2 << endl; else if (AMOUNT >= 21 && AMOUNT <= 40) cout << 80 + (AMOUNT - 20) << endl; else if (AMOUNT > 40) cout << 100 << endl; } }
代码如上;
不难,只是用多重if语句;
AC的,没错。




#17261: Re:C++ CPP


ufve0704 (爬 我爬 我爬爬爬 有排行榜這種東西就是要爬 爬過我上面的那...)


#include <bits/stdc++.h>

using namespace std;

int a; 

int main(int argc, char** argv){

while(cin>>a)

cout<<(a<=10)*(a*6)+(a>=11&&a<=20)*((a-10)*2+60)+(a>=21&&a<40)*((a-20)+80)+(a>=40)*100<<endl;

}

無聊來嘗試cout單行解