AcWing 669. 加薪
原题链接
简单
作者:
mmmorning
,
2024-04-12 21:24:16
,
所有人可见
,
阅读 2
(循环)简化判断输出
C++ 代码
感谢观看。个人理解,如有问题,还请批评指正。
#include <iostream>
#include <cstdio>
using namespace std;
int main(){
double x;
cin >> x;
int a[]={400,800,1200,2000,15,12,10,7,4};
for(int i=0;i<4;i++){
if(i==3&&x>2000){
printf("Novo salario: %.2lf\n",x*(1+a[i+5]*0.01));
printf("Reajuste ganho: %.2lf\n",x*a[i+5]*0.01);
printf("Em percentual: %d %\n",a[i+5]);
}
if(x<=a[i]){
printf("Novo salario: %.2lf\n",x*(1+a[i+4]*0.01));
printf("Reajuste ganho: %.2lf\n",x*a[i+4]*0.01);
printf("Em percentual: %d %\n",a[i+4]);
break; //跳出循环。已有符合条件的输出,不用再进行循环判断
}
/*else if(x>a[i]){
continue;
}*/
}
return 0;
}