#include<iostream> #include<cmath> using namespace std; int main(){ int n; cin>>n; int count=0; //总短信数 double cost=0; for(int i=0,x;i<n;i++){ cin>>x; count+=ceil((double)x/70); //向上取整 } cost=count*0.1; cout<<cost<<endl; return 0; }

答案

那肯定得用啊 题目叫我用我偏不用](https://www.acgo.cn/problemset/423/explanation)

#include<bits/stdc++.h> using namespace std; int main(){ int a; cin>>a; double f=0.0; int b[a+1]; for(int i=1;i<=a;i++){ cin>>b[i]; if(b[i]>70){ if(b[i]%70==0){ f+=b[i]/70*1; } else f+=b[i]/70+1; } else f+=1; } cout<<f/10; return 0; }

#include <iostream> using namespace std; int main (){ int n; int sum=0; int x[10001]; cin >> n; for(int i=0;i<=n;i++){ cin >> x[i]; }for(int i=0;i<=n;i++){ if(x[i]>70&&x[i]%70==0){ sum=sum+x[i]/70; }else{ if(x[i]<=70){ sum=sum+1; }else{ if(x[i]>70&&x[i]%70!=0){ sum=sum+x[i]/70+1; } } } } cout << sum*0.1-0.1; return 0; }

#include<bits/stdc++.h> using namespace std; int main() { int n; cin>>n; int sum=0 ; double k=0 ; for(int i=0 ,x; i<n; i++) { cin>>x; sum+= ceil((double)x/70 ); } k=sum* 0.1 ; cout<<k<<endl; return 0 ; }

我的帖子首页

#include<iostream> using namespace std; int main() { int n; double sum = 0.0; cin >> n; for(int i = 1; i <= n; i++) { int x; cin >> x; if(x <= 70) { sum += 0.1; }else{ sum += 0.1+(x/70)*0.1; } } cout << sum; return 0; }

就 不给答案 略略略 骗你的 正确答案如下

#include<iostream> #include<cmath> using namespace std; int main(){ int n; cin>>n; int count=0; //总短信数 double cost=0; for(int i=0,x;i<n;i++){ cin>>x; count+=ceil((double)x/70); //向上取整 } cost=count*0.1; cout<<cost<<endl; return 0; }

#include <iostream> #include <iomanip> // 用于设置输出精度 using namespace std; int main() { int n; cin >> n; // 读取短信发送次数 }