vector_STL_ @ 2024-08-19 18:55:38
#include<bits/stdc++.h>
using namespace std;
int main(){
long long a[15];
cin>>a[1]>>a[2]>>a[3]>>a[4]>>a[5]>>a[6]>>a[7]>>a[8]>>a[9]>>a[10];
cout<<a[1]*28.9+a[2]*32.7+a[3]*45.6+a[4]*78+a[5]*35+a[6]*86.2+a[7]*27.8+a[8]*43+a[9]*56+a[10]*65;
return 0;
}
by gongziwen @ 2024-08-19 18:57:15
@zch120526 《输出格式》《包含一个实数》《精确到小数点后一位》
by zhizhenyaohanyu @ 2024-08-19 19:00:10
要保留小数点后1位,下次记得看输出格式@zch120526
by sean0613 @ 2024-08-19 19:10:08
@zch120526 《后一位》
by BlackWuKong @ 2024-08-19 19:10:36
@zch120526 我的思路:
#include<bits/stdc++.h>
using namespace std;
int main(){
double p[110]={0,28.9,32.7,45.6,78,35,86.2,27.8,43,56,65},m=0;
int a;
for (int i=1;i<=10;i++){
cin>>a;
m+=a*p[i];
}
printf("%.1lf",m);
return 0;
}
求关
by lce11451410086 @ 2024-08-23 13:32:33
@zch120526 我的思路
#include<bits/stdc++.h>
using namespace std;
int main(){
double a[10]={28.9,32.7,45.6,78,35,86.2,27.8,43,56,65};
double sum=0;
for(auto p:a){
int x;
cin>>x;
sum+=x*p;
}
printf("%.1lf",sum);
return 0;
}
by chenzhishuo2012 @ 2024-09-05 07:34:11
@zch120526 求互关
#include<bits/stdc++.h>
using namespace std;
int a[20];
double b[20]={0,28.9,32.7,45.6,78,35,86.2,27.8,43,56,65},ans;
int main(){
for(int i=1;i<=10;i++){
cin>>a[i];
ans+=a[i]*b[i];
}
printf("%0.1lf",ans);
return 0;
}
by zycEthan @ 2024-12-04 20:16:47
我的思路:
#include<bits/stdc++.h>
using namespace std;
int main(){
float a[10]={28.9,32.7,45.6,78,35,86.2,27.8,43,56,65},x=0;
int i,n;
for (i=0;i<10;i++){
cin >> n;
x+=n*a[i];
}
cout << x*1.0;
}
}