1104. Sum of Number Segments (20)
2016-11-29 22:47
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AC代码
#include<stdio.h>
#include<vector>
using namespace std;
vector<double> a;
int main(){
int n;
scanf("%d",&n);
double sum=0;
for(int i=0;i<n;i++){
double temp;
scanf("%lf",&temp);
a.push_back(temp);
}
for(int i=0;i<n;i++){
sum+=a[i]*(n-i)*(i+1);
}
printf("%.2lf\n",sum);
}
原先写的,和上面的应该一样,但就是一个测试点不过
#include<stdio.h>
#include<vector>
using namespace std;
vector<double> a;
int main(){
int n;
scanf("%d",&n);
double sum=0;
for(int i=0;i<n;i++){
double temp;
scanf("%lf",&temp);
a.push_back(temp);
}
for(int i=0;i<n/2;i++){
sum+=(a[i]+a[n-1-i])*(n-i)*(i+1);
}
if(n%2==1){
sum+=a[n/2]*(n-n/2)*(n/2+1);
}
printf("%.2lf\n",sum);
}
#include<stdio.h>
#include<vector>
using namespace std;
vector<double> a;
int main(){
int n;
scanf("%d",&n);
double sum=0;
for(int i=0;i<n;i++){
double temp;
scanf("%lf",&temp);
a.push_back(temp);
}
for(int i=0;i<n;i++){
sum+=a[i]*(n-i)*(i+1);
}
printf("%.2lf\n",sum);
}
原先写的,和上面的应该一样,但就是一个测试点不过
#include<stdio.h>
#include<vector>
using namespace std;
vector<double> a;
int main(){
int n;
scanf("%d",&n);
double sum=0;
for(int i=0;i<n;i++){
double temp;
scanf("%lf",&temp);
a.push_back(temp);
}
for(int i=0;i<n/2;i++){
sum+=(a[i]+a[n-1-i])*(n-i)*(i+1);
}
if(n%2==1){
sum+=a[n/2]*(n-n/2)*(n/2+1);
}
printf("%.2lf\n",sum);
}
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