1 solutions
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0
C :
#include <stdio.h> #include <math.h> #include <string.h> int main(){ int i,n; scanf ("%d",&n); float a[100]; for (i=0;i<n;i++) scanf ("%f",&a[i]); float max=a[0],min=a[0]; for (i=0;i<n;i++) { max=max>a[i]?max:a[i]; min=min>a[i]?a[i]:min; } printf ("%.2f %.2f\n",max,min); return 0; }C++ :
#include<iostream> #include<cstdio> using namespace std; float a[10000]; float fmax(float a[],int n) //查找最大值 { float max=a[1]; for (int i=2; i<=n; i++) if (max<a[i]) max=a[i]; return max; } float fmin(float a[],int n) //查找最小值 { float min=a[1]; for (int i=2; i<=n; i++) if (min>a[i]) min=a[i]; return min; } int main() { int n; cin>>n; for (int i=1; i<=n; i++) cin>>a[i]; printf("%.2f %.2f\n",fmax(a,n),fmin(a,n)); return 0; }
- 1
Information
- ID
- 17563
- Time
- 1000ms
- Memory
- 128MiB
- Difficulty
- (None)
- Tags
- # Submissions
- 0
- Accepted
- 0
- Uploaded By