。。。double类型无法通过

绕钉子的长绳子
wsadlxx LV 8 @ 2016-09-07 19:22:51

float就好。

（语言： c）

twd2 LV 9 MOD @ 2016-09-10 12:08:56

http://stackoverflow.com/questions/4264127/correct-format-specifier-for-double-in-printf
liuyiah LV 10 @ 2016-09-10 11:34:21

实际上printf用%f既可以输出float也可输出double
Heartbeat LV 10 @ 2016-09-07 22:55:39

可以啊为什么不行
- wsadlxx LV 8 @ 2016-09-07 23:32:28
  
  不知道啊，我用同样的程序，double的话就是WA，float的话就AC。。。
- lrj124 LV 10 @ 2016-09-08 13:19:40
  
  ###****double AC****
```c++
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 504 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 508 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 508 KiB, score = 10
Accepted, time = 0 ms, mem = 508 KiB, score = 60
代码
#include<cstdio>
#include<cmath>
using namespace std;
int main()
{
int n;
double a[101][3],ans=0,r;
scanf("%d %lf\n",&n,&r);
for(int i=1;i<=n;i++)
scanf("%lf %lf",&a[i][1],&a[i][2]);
for(int i=1;i<n;i++)
ans+=sqrt((a[i][1]-a[i+1][1])*(a[i][1]-a[i+1][1])+(a[i][2]-a[i+1][2])*(a[i][2]-a[i+1][2]));
ans+=sqrt((a[n][1]-a[1][1])*(a[n][1]-a[1][1])+(a[n][2]-a[1][2])*(a[n][2]-a[1][2]));
printf("%.2lf",ans+r*2*3.1415926);
return 0;
}
```
- wsadlxx LV 8 @ 2016-09-08 14:23:13
  
  这就尴尬了。。。
  这是为什么呢？。。。
  评测结果
  编译成功
  
  测试数据 #0: WrongAnswer, time = 0 ms, mem = 488 KiB, score = 0
  测试数据 #1: WrongAnswer, time = 0 ms, mem = 484 KiB, score = 0
  测试数据 #2: WrongAnswer, time = 0 ms, mem = 484 KiB, score = 0
  测试数据 #3: WrongAnswer, time = 0 ms, mem = 484 KiB, score = 0
  测试数据 #4: WrongAnswer, time = 0 ms, mem = 484 KiB, score = 0
  测试数据 #5: WrongAnswer, time = 0 ms, mem = 488 KiB, score = 0
  WrongAnswer, time = 0 ms, mem = 488 KiB, score = 0
  代码
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <malloc.h>
#include <math.h>
#define pi 3.1415926
int n;
double r;
double result;
struct point
{
double x,y;
}p0,p1,p2;
int i;

double dis(struct point p1, struct point p2)
{
double result;
result = sqrt(pow(p1.x-p2.x,2) + pow(p1.y-p2.y,2));
return result;
}
int main()
{
result=0.0;
scanf("%d %lf",&n,&r);
scanf("%lf %lf",&p0.x, &p0.y);
p1=p0;
for (i=1;i<n;i++)
{
scanf("%lf %lf",&p2.x,&p2.y);
result+=dis(p1,p2);
p1=p2;
}
result+=dis(p1,p0);
result+=2*pi*r;
printf("%.2lf",result);

return 0;
}
```
- lrj124 LV 10 @ 2016-09-09 13:44:20
  
  你的代码明明是AC的
  评测结果
  编译成功
  
  测试数据 #0: Accepted, time = 0 ms, mem = 508 KiB, score = 10
  测试数据 #1: Accepted, time = 0 ms, mem = 508 KiB, score = 10
  测试数据 #2: Accepted, time = 0 ms, mem = 508 KiB, score = 10
  测试数据 #3: Accepted, time = 0 ms, mem = 508 KiB, score = 10
  测试数据 #4: Accepted, time = 0 ms, mem = 508 KiB, score = 10
  测试数据 #5: Accepted, time = 0 ms, mem = 504 KiB, score = 10
  Accepted, time = 0 ms, mem = 508 KiB, score = 60
  代码
```c
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <malloc.h>
#include <math.h>
#define pi 3.1415926
int n;
double r;
double result;
struct point
{
double x,y;
}p0,p1,p2;
int i;

double dis(struct point p1, struct point p2)
{
double result;
result = sqrt(pow(p1.x-p2.x,2) + pow(p1.y-p2.y,2));
return result;
}
int main()
{
result=0.0;
scanf("%d %lf",&n,&r);
scanf("%lf %lf",&p0.x, &p0.y);
p1=p0;
for (i=1;i<n;i++)
{
scanf("%lf %lf",&p2.x,&p2.y);
result+=dis(p1,p2);
p1=p2;
}
result+=dis(p1,p0);
result+=2*pi*r;
printf("%.2lf",result);

return 0;
}
```
- wsadlxx LV 8 @ 2016-09-09 15:05:57
  
  你选择的编译语言是 C++
  我选择的编译语言是 C
  。。。
  我不知道为什么C++能过，C不能过。
  大神求破~
- lrj124 LV 10 @ 2016-09-09 15:17:36
  
  ###****c语言double输入是%lf，输出是%f****
```c++
记录信息
评测状态 Accepted
题目 P1007 绕钉子的长绳子
递交时间 2016-09-09 15:16:07
代码语言 C
评测机 ShadowShore
消耗时间 0 ms
消耗内存 488 KiB
评测时间 2016-09-09 15:16:07
评测结果
编译成功

测试数据 #0: Accepted, time = 0 ms, mem = 484 KiB, score = 10
测试数据 #1: Accepted, time = 0 ms, mem = 484 KiB, score = 10
测试数据 #2: Accepted, time = 0 ms, mem = 484 KiB, score = 10
测试数据 #3: Accepted, time = 0 ms, mem = 484 KiB, score = 10
测试数据 #4: Accepted, time = 0 ms, mem = 484 KiB, score = 10
测试数据 #5: Accepted, time = 0 ms, mem = 488 KiB, score = 10
Accepted, time = 0 ms, mem = 488 KiB, score = 60
代码
#include <stdio.h>
#include <stdlib.h>
#include <string.h>
#include <malloc.h>
#include <math.h>
#define pi 3.1415926
int n;
double r;
double result;
struct point
{
double x,y;
}p0,p1,p2;
int i;

double dis(struct point p1, struct point p2)
{
double result;
result = sqrt(pow(p1.x-p2.x,2) + pow(p1.y-p2.y,2));
return result;
}
int main()
{
result=0.0;
scanf("%d %lf",&n,&r);
scanf("%lf %lf",&p0.x, &p0.y);
p1=p0;
for (i=1;i<n;i++)
{
scanf("%lf %lf",&p2.x,&p2.y);
result+=dis(p1,p2);
p1=p2;
}
result+=dis(p1,p0);
result+=2*pi*r;
printf("%.2f",result);

return 0;
}
```
- wsadlxx LV 8 @ 2016-09-09 20:05:34
  
  哦哦~ 学习了~ 谢谢~