大家看看这道题，我感觉我没有做错!为什么总是wrong anwser 呢！
Number sequence
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Time limit: 1sec. Submitted: 31
Memory limit: 64M Accepted: 8
Source : SCU Programming Contest 2006 Final
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Given a number sequence which has N element(s), please calculate the number of different collocation for three number Ai, Aj, Ak, which satisfy that Ai < Aj > Ak and i < j < k.
Input
The first line is an integer N (N <= 50000). The second line contains N integer(s): A1, A2, ..., An(0 <= Ai <= 32768).
Output
There is only one number, which is the the number of different collocation.
Sample Input
5
1 2 3 4 1
Sample Output
6
#include <stdio.h>
#define N 1000
int main(void)
{
long s;
int str1[N], i, j, k = 1, num = 0;
scanf("%ld", &s);
for(i = 0;i < s;i ++)
scanf("%d", &str1[i]);
for(k = 1;k < s - 1;k ++)
for(i = 0;i < k;i ++)
for(j = k + 1;j < s;j ++)
{
if(str1[k] > str1[i]&&str1[k] >str1[j])
num ++;
}
printf("%d\n", num);
return 0;
}
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有谁能翻译一下?
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没看你的程序,但一点你就错了
N (N <= 50000). The second line contains N integer(s): A1, A2, ..., An(0 <= Ai <= 32768).
而你定义的数组才多大.
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楼上,lz说总是错误,一定还有其它原因的
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现在去吃饭,所以没看程序,等回来再帮LZ看看
不好意思哈.
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我数组也定义过很大！但是还是错误！
这道题的大概意思是先输入一个数字，是一个数列当中元素的个数！输入整个数列之后！判断这个数列中的三个数，要求其中中间的数要大于两边的数！Ai < Aj > Ak i < j < k这三个数还要求是按原数列当中的顺序排列！
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呵呵！没有关系！
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#include <stdio.h>
#define N 4000
int main(void)
{
long s;
int str1[N]={0},i,j,k;
long num=0;
scanf("%ld", &s);
for(i=0L;i<s;i++)
scanf("%d",&str1[i]);
for(k=1;k<s-1;k++)
for(i=0;i<k;i++)
for(j=k+1;j<s;j++)
if(str1[k]>str1[i] && str1[k]>str1[j]) num++;
printf("%ld\n",num);
getch();
return 0;
}
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我交了，还是错误！
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#include<stdio.h>
int main()
{
int i,j,k,n;
unsigned int a[50000];
long count=0;
scanf("%d",&n);
for(i=0;i<n;i++)
{
scanf("%d",&a[i]);
}
for(i=0;i<n-2;i++)
{
for(j=i+1;j<n-1;j++)
{
for(k=j+1;k<n;k++)
{
if(a[i]<a[j]&&a[j]>a[k])
{
count++;
}
}
}
}
printf("%ld\n",count);
return (0);
}
/*估计会超时*/
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