2 sakura miss Sakura_miss 于 2018.02.07 19:20 提问

c++提交了很多次,但是都WA了,检查了几遍,但是总不知道错在哪里了,求解。万分感谢
c++

D - Average Sleep Time

It's been almost a week since Polycarp couldn't get rid of insomnia. And as you may already know, one week in Berland lasts k days!

When Polycarp went to a doctor with his problem, the doctor asked him about his sleeping schedule (more specifically, the average amount of hours of sleep per week). Luckily, Polycarp kept records of sleep times for the last n days. So now he has a sequence a1, a2, ..., an, where ai is the sleep time on the i-th day.

The number of records is so large that Polycarp is unable to calculate the average value by himself. Thus he is asking you to help him with the calculations. To get the average Polycarp is going to consider k consecutive days as a week. So there will be n - k + 1 weeks to take into consideration. For example, if k = 2, n = 3 and a = [3, 4, 7], then the result is .

You should write a program which will calculate average sleep times of Polycarp over all weeks.

Input
The first line contains two integer numbers n and k (1 ≤ k ≤ n ≤ 2·105).

The second line contains n integer numbers a1, a2, ..., an (1 ≤ ai ≤ 105).

Output
Output average sleeping time over all weeks.

The answer is considered to be correct if its absolute or relative error does not exceed 10 - 6. In particular, it is enough to output real number with at least 6 digits after the decimal point.

Example
Input
3 2
3 4 7
Output
9.0000000000
Input
1 1
10
Output
10.0000000000
Input
8 2
1 2 4 100000 123 456 789 1
Output
28964.2857142857

我的代码:
#include
#include
#include
#include
#include
using namespace std;
#define N 200005
long long int a[N];
int main()
{
long long int n,k,i,j;
double sum = 0,sum1 = 0;
scanf("%lld %lld",&n,&k);
for(int i = 0;i < n; i++)
scanf("%lld",&a[i]);
if(n == k)
{
for(int i = 0;i < n; i++)
{
sum += a[i];
}
printf("%.10lf\n",sum);
}
else
{
j = 0;
for(i = 1;i <= n - k; i++) //n - k表示当有n - k + 1个周时,会有n - k个重复加的数字。
{
j += k - 1;
sum1 += 2 * a[j];
a[j] = 0;
}
for(int i = 0;i < n; i++)
{
if(a[i] != 0)
sum1 += a[i];
}
double ans = sum1 /((n - k + 1) * 1.0);
printf("%.10lf\n",ans);
}
return 0;
}

6个回答

qq_18502471
qq_18502471   2018.02.07 20:30
已采纳

你对题目理解有问题
比如说
7 3
1 2 3 4 5 6 7
应该这样算
图片说明
但用你的代码肯定不会,你看你代码sum1 += 2 * a[j]; 很明显我这个3到5每个数字都加了3次

MDZZ英文题目最讨嫌,还是百度答案把

Sakura_miss
Sakura_miss 是我理解错题意了。
15 天之前 回复
ffyy88jh
ffyy88jh   2018.02.07 20:56

#define N 200005
long long int a[N];
int main()
{
long long int n,k,i,j;
double sum;
while (scanf("%lld %lld",&n,&k)>1) //ctrl+z 结束输入
{
for(i = 0; i < n; i++)
scanf("%lld",&a[i]);
for(sum=i=0; i<n-k+1; i++)
{ for(j=0; j<k; j++)
sum += a[i+j];
}
printf("%.10lf\n",sum/(n - k + 1));
}
return 0;
}

hfhbutn
hfhbutn   2018.02.07 19:40

i重复定义很多次了吧,为啥不换一个名字的变量。

Sakura_miss
Sakura_miss 但不影响结果呀,我下次会注意这点,但为啥就是AC不了,我觉得思路没错呀
15 天之前 回复
ffyy88jh
ffyy88jh   2018.02.07 21:14

//如果每次计算一组,可以简化为:
#define N 200005
long int a[N];
int main()
{
long int n,k,i,j;
double sum;
scanf ("%ld %ld",&n,&k);
for ( i = 0; i < n; i++)
scanf ("%ld",&a[i]);
for ( sum=i=0; i<n-k+1; i++)
for ( j=0; j<k; j++)
sum += a[i+j];
printf("%.10lf\n",sum/(n - k + 1));
return 0;
}

Sakura_miss
Sakura_miss 谢谢你的回答,很棒哟。才发现可以用嵌套FOR循环来解决。
15 天之前 回复
qq_37233607
qq_37233607   2018.02.07 20:23

你好,你的代码我在VS2013上测试过.没有任何问题,唯一不同的地方在于,我在最前面加上了#pragma warning (disable : 4996) 这个命令吧.毕竟scanf不安全.要么把scanf改为scanf_s,要么就加上#pragma warning (disable : 4996)

qq_19688999
qq_19688999 #define N 200005 long long int a[N]; int main() { long long int n,k,i,j; double sum; while (scanf("%lld %lld",&n,&k)>1) //ctrl+z 结束输入 { for(i = 0; i < n; i++) scanf("%lld",&a[i]); for(sum=i=0; i<n-k+1; i++) { for(j=0; j<k; j++) sum += a[i+j]; } printf("%.10lf\n",sum/(n - k + 1)); } return 0; }
15 天之前 回复
Sakura_miss
Sakura_miss 我在编译器上也试过题目所给的样例,都能运行出来,但是VJ上面就过不去,不知道是不是思路错了。可是找不出自己的问题,很心塞,求指教。
15 天之前 回复
qq_26103095
qq_26103095   2018.02.07 21:38

//如果每次计算一组,可以简化为:
#define N 200005
long int a[N];
int main()
{
long int n,k,i,j;
double sum;
scanf ("%ld %ld",&n,&k);
for ( i = 0; i < n; i++)
scanf ("%ld",&a[i]);
for ( sum=i=0; i<n-k+1; i++)
for ( j=0; j<k; j++)
sum += a[i+j];
printf("%.10lf\n",sum/(n - k + 1));
return 0;
}

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